寒假第二周 1.18 --- 1.24
1.18
cf 581c
581C - Developing Skills
重新自己写了一遍，注意都是0 的时候
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 1e5+5; 9 int n,k; 10 11 struct node{ 12 int x,y; 13 }p[maxn]; 14 15 int cmp(node n1,node n2){ 16 return n1.y < n2.y; 17 } 18 19 int cmp0(node n1,node n2){ 20 return n1.x < n2.x; 21 } 22 23 int a[105]; 24 25 void solve(){ 26 for(int i = 1;i <= n;i++){ 27 int pos = lower_bound(a+1,a+10,p[i].x) - a; 28 if(a[pos] == p[i].x) pos++; 29 p[i].y = a[pos] - p[i].x; 30 // printf("a[%d] = %d p[%d].x = %d\n",pos,a[pos],i,p[i].x); 31 } 32 /* for(int i = 1;i <= n;i++){ 33 printf("p[%d].x = %d y = %d\n",i,p[i].x,p[i].y); 34 }*/ 35 sort(p+1,p+n+1,cmp); 36 int ans = 0; 37 for(int i = 1;i <= n;i++){ 38 if(k >= p[i].y){ 39 ans += (p[i].x + p[i].y)/10; 40 k -= p[i].y; 41 p[i].x = p[i].x + p[i].y; 42 } 43 else { 44 ans += p[i].x/10; 45 } 46 // printf("i = %d k = %d ans = %d\n",i,k,ans); 47 } 48 if(k){ 49 for(int i = 1;i <= n;i++){ 50 int l = 10 - p[i].x/10; 51 int r = k/10; 52 if(r >= l){ 53 ans += l; 54 k = k-l*10; 55 } 56 else{ 57 ans += k/10; 58 k = k%10; 59 } 60 if(k < 10) break; 61 //printf("l = %d r = %d k = %\n",l,r,k); 62 } 63 } 64 printf("%d\n",ans); 65 } 66 67 int main(){ 68 for(int i = 1;i <= 10;i++) a[i] = i*10; 69 a[11] = 100; 70 // freopen("in.txt","r",stdin); 71 // freopen("out.txt","w",stdout); 72 while(scanf("%d %d",&n,&k) != EOF){ 73 for(int i = 1;i <= n;i++){ 74 scanf("%d",&p[i].x); 75 } 76 solve(); 77 } 78 return 0; 79 }
cf 614e
614E - Necklace
先不理解题解说的 part 是什么意思，，就是所有a[i] 的gcd
然后像题解说的那样构造
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 1e5+5; 9 char s[maxn]; 10 int n,a[maxn],aa[maxn]; 11 12 int gcd(int a,int b){ 13 return (!b) ? a:gcd(b,a%b); 14 } 15 16 void solve(){ 17 int part = a[1]; 18 for(int i = 2;i <= n;i++){ 19 part = gcd(part,a[i]); 20 } 21 for(int i = 1;i <= n;i++){ 22 aa[i] = a[i]/part; 23 } 24 25 if(part%2 == 0){ 26 printf("%d\n",part); 27 vector<char> c; 28 for(int i = 1;i <= n;i++){ 29 for(int j = 1;j <= aa[i];j++){ 30 c.push_back(i+'a'-1); 31 } 32 } 33 int flag = 1; 34 for(int i = 1;i <= part;i++){ 35 if(flag){ 36 for(int j = 0;j < c.size();j++){ 37 printf("%c",c[j]); 38 } 39 } 40 else{ 41 for(int j = c.size()-1;j >= 0;j--){ 42 printf("%c",c[j]); 43 } 44 } 45 flag = !flag; 46 } 47 printf("\n"); 48 return; 49 } 50 int ji = 0; 51 for(int i = 1;i <= n;i++){ 52 ji += (aa[i]%2); 53 } 54 if(ji > 1){ 55 puts("0"); 56 for(int i = 1;i <= n;i++){ 57 for(int j = 1;j <= a[i];j++){ 58 char z = i+'a'-1; 59 printf("%c",z); 60 } 61 } 62 printf("\n"); 63 } 64 if(ji == 1){ 65 vector<char> l; 66 vector<char> r; 67 int pos; 68 for(int i = 1;i <= n;i++){ 69 if(a[i]%2){ 70 pos = i; 71 break; 72 } 73 } 74 for(int i = 1;i <= n;i++){ 75 if(i == pos) continue; 76 for(int j = 1;j <= aa[i]/2;j++){ 77 l.push_back(i+'a'-1); 78 r.push_back(i+'a'-1); 79 } 80 } 81 for(int j = 1;j <= aa[pos];j++){ 82 l.push_back(pos+'a'-1); 83 } 84 reverse(r.begin(),r.end()); 85 for(int j = 0;j < r.size();j++){ 86 l.push_back(r[j]); 87 } 88 printf("%d\n",part); 89 for(int i = 1;i <= part;i++){ 90 for(int j = 0;j < l.size();j++){ 91 printf("%c",l[j]); 92 } 93 } 94 printf("\n"); 95 } 96 } 97 98 int main(){ 99 while(scanf("%d",&n) != EOF){ 100 for(int i = 1;i <= n;i++){ 101 scanf("%d",&a[i]); 102 } 103 solve(); 104 } 105 return 0; 106 }
hdu 1069
加了一维的 dag 上的动规，去年的时候就在紫薯上看，不会写- -
然后，sb的我是这样的，用 g[i][j][k][z] 建图，表示 第 i 个立方体 的第k 面为底面的时候，能够放在 第 j 个立方体 的第z面为底面的时候
后来不会写了---
-------------------
稍微转化一下就和普通的一样了，就是每个立方体三种放置的情况分别拆成三个立方体，就可以了
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 1005; 9 int dp[maxn],n; 10 11 struct node{ 12 int x,y,z; 13 }p[maxn]; 14 15 int dfs(int i){ 16 if(dp[i]) return dp[i]; 17 int ans = 0; 18 for(int j = 1;j <= 3*n;j++){ 19 if(p[j].x > p[i].x && p[j].y > p[i].y){ 20 ans = max(ans,dfs(j)); 21 } 22 } 23 return dp[i] = ans + p[i].z; 24 } 25 26 int main(){ 27 int kase = 0; 28 while(scanf("%d",&n) != EOF){ 29 if(n == 0) break; 30 int a[5]; 31 int cnt = 0; 32 for(int i = 1;i <= n;i++){ 33 scanf("%d %d %d",&a[0],&a[1],&a[2]); 34 sort(a,a+3); 35 p[++cnt].x = a[0];p[cnt].y = a[1];p[cnt].z = a[2]; 36 p[++cnt].x = a[1];p[cnt].y = a[2];p[cnt].z = a[0]; 37 p[++cnt].x = a[0];p[cnt].y = a[2];p[cnt].z = a[1]; 38 } 39 int res = -1; 40 memset(dp,0,sizeof(dp)); 41 for(int i = 1;i <= 3*n;i++){ 42 res = max(res,dfs(i)); 43 } 44 printf("Case %d: maximum height = %d\n",++kase,res); 45 } 46 return 0; 47 }
hdu 1087
最大的严格上升的序列的和
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 typedef long long LL; 8 const int maxn = 1005; 9 int n,a[maxn]; 10 LL dp[maxn]; 11 12 void solve(){ 13 memset(dp,0,sizeof(dp)); 14 a[0] = 0; 15 for(int i = 1;i <= n;i++){ 16 for(int j = 1;j < i;j++){ 17 if(a[i] > a[j]){ 18 dp[i] = max(dp[i],dp[j]+1LL*a[i]); 19 } 20 } 21 dp[i] = max(dp[i],1LL*a[i]); 22 // printf("dp[%d] = %d\n",i,dp[i]); 23 } 24 LL ans = -1; 25 for(int i = 1;i <= n;i++){ 26 ans = max(ans,dp[i]); 27 } 28 printf("%I64d\n",ans); 29 } 30 31 int main(){ 32 while(scanf("%d",&n) != EOF){ 33 if(n == 0) break; 34 for(int i = 1;i <= n;i++){ 35 scanf("%d",&a[i]); 36 } 37 solve(); 38 } 39 return 0; 40 } 41 42 FAQ | About Virtual Judge | Forum | Discuss | Open Source Project 43 All Copyright Reserved ©2010-2014 HUST ACM/ICPC TEAM 44 Anything about the OJ, please ask in the forum, or contact author:Isun 45 Server Time: 2016-01-19 22:29:09
hdu 1114
完全背包必须装满的最优解
dp[0] 初始化 为 0 ，其余容量初始化为 无穷大
---话说就搜了下完全背包必须装满，题解都出来了-----
，只有初始容量为0价值为 0 的包装满才是合法的解，
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 typedef long long LL; 8 const int INF = (1<<30)-1; 9 const int maxn = 10005; 10 int w[maxn],v[maxn],n,e,f,V; 11 int dp[maxn]; 12 13 void solve(){ 14 dp[0] = 0; 15 V = f-e; 16 for(int i = 1;i <= V;i++) dp[i] = INF; 17 for(int i = 1;i <= n;i++){ 18 for(int j = w[i];j <= V;j++){ 19 dp[j] = min(dp[j-w[i]]+v[i],dp[j]); 20 // printf("dp[%d] = %d\n",j,dp[j]); 21 } 22 } 23 if(dp[V] == INF) puts("This is impossible."); 24 else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]); 25 } 26 27 int main(){ 28 int T; 29 scanf("%d",&T); 30 while(T--){ 31 scanf("%d %d",&e,&f); 32 scanf("%d",&n); 33 for(int i = 1;i <= n;i++){ 34 scanf("%d %d",&v[i],&w[i]); 35 } 36 solve(); 37 } 38 return 0; 39 }
1.19
hdu 1176
数塔
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 1e5+5; 8 int dp[maxn][15],a[maxn][15]; 9 int n,T; 10 11 void solve(){ 12 memset(dp,0,sizeof(dp)); 13 for(int i = T;i >= 0;i--){ 14 dp[i][0] = max(dp[i+1][0],dp[i+1][1]) + a[i][0]; 15 // printf("dp[%d][0] = %d\n",i,dp[i][0]); 16 for(int j = 1;j <= 9;j++){ 17 dp[i][j] = max(max(dp[i+1][j-1],dp[i+1][j+1]),dp[i+1][j]) + a[i][j]; 18 // printf("dp[%d][%d] = %d\n",i,j,dp[i][j]); 19 } 20 dp[i][10] = max(dp[i+1][10],dp[i+1][9]) + a[i][10]; 21 // printf("dp[%d][10] = %d\n",i,dp[i][10]); 22 } 23 printf("%d\n",dp[0][5]); 24 } 25 26 int main(){ 27 while(scanf("%d",&n) != EOF){ 28 if(n == 0) break; 29 memset(a,0,sizeof(a)); 30 int x,t; 31 T = -1; 32 for(int i = 1;i <= n;i++){ 33 scanf("%d %d",&x,&t); 34 a[t][x]++; 35 T = max(T,t); 36 } 37 solve(); 38 } 39 return 0; 40 }
hdu 1260
...燃烧生命wa 水题.....
输出的时间最大是 12
-----一直纳闷---我有%12阿
-------
12:00:00 是 am（还要再判断下分钟和秒钟---sad---）
然后再多一点点就是 pm 了
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 5005; 9 int dp[maxn],a[maxn],b[maxn]; 10 int n; 11 const int INF = (1<<30)-1; 12 13 void solve(){ 14 for(int i = 1;i < maxn;i++) dp[i] = INF; 15 dp[0] = 0; 16 for(int i = 1;i <= n;i++){ 17 if(i == 1){ 18 dp[i] = a[i]; 19 } 20 if(i >= 2){ 21 dp[i] = min(dp[i-1]+a[i],dp[i-2]+b[i-1]); 22 } 23 // printf("dp[%d] = %d\n",i,dp[i]); 24 } 25 int shi = dp[n]/3600; 26 int fen = (dp[n]-shi*3600)/60; 27 int miao = dp[n] - shi*3600-fen*60; 28 char c1,c2; 29 shi = shi+8; 30 if(shi < 12||shi == 12 && fen == 0 && miao == 0){ 31 c1 = 'a'; 32 c2 = 'm'; 33 } 34 else{ 35 if(shi != 12) shi %= 12; 36 c1 = 'p'; 37 c2 = 'm'; 38 } 39 // printf("shi = %d fen = %d miao = %d\n",shi,fen,miao); 40 printf("%02d:%02d:%02d %c%c\n",shi,fen,miao,c1,c2); 41 } 42 43 int main(){ 44 int T; 45 scanf("%d",&T); 46 while(T--){ 47 scanf("%d",&n); 48 for(int i = 1;i <= n;i++) scanf("%d",&a[i]); 49 for(int i = 1;i <= n-1;i++) scanf("%d",&b[i]); 50 solve(); 51 } 52 return 0; 53 }
cf 602 d 602D - Lipshitz Sequence
b[i] = abs(a[i] - a[i-1]) ，求出每个b[i] 作为最大值的时候，向左能延伸多远，向右能延伸多远
---话说是看的题解补的题---
然后因为两个地方没有注意到，代码一直不对，
就是比如现在的询问是 [l,r]
b[l] 是 a[l] 与 a[l-1] 的差值，不包含在这个区间里面，所以应该 l++;
还有就是递推 left[]，right[] 数组的时候，只需要一个是 >= ，免得把区间算重了--
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 1e6+5; 9 typedef long long LL; 10 int n,l[maxn],r[maxn]; 11 int a[maxn],b[maxn]; 12 int q; 13 14 void solve(){ 15 memset(b,0,sizeof(b)); 16 memset(l,0,sizeof(l)); 17 memset(r,0,sizeof(r)); 18 for(int i = 2;i <= n;i++){ 19 b[i] = abs(a[i] - a[i-1]); 20 } 21 22 /* for(int i = 2;i <= n;i++){ 23 printf("%d ",b[i]); 24 } 25 printf("\n");*/ 26 27 for(int i = 1;i <= q;i++){ 28 int lb,ub; 29 scanf("%d %d",&lb,&ub); 30 lb++; 31 for(int j = lb;j <= ub;j++){ 32 l[j] = j; 33 r[j] = j; 34 } 35 for(int j = lb;j <= ub;j++){ 36 while(l[j] > lb && b[j] > b[l[j]-1]){ 37 l[j] = l[l[j]-1]; 38 } 39 } 40 for(int j = ub;j >= lb;j--){ 41 while(r[j] < ub && b[j] >= b[r[j]+1]){ 42 r[j] = r[r[j]+1]; 43 } 44 } 45 LL ans = 0; 46 for(int j = lb;j <= ub;j++){ 47 LL cnt = (1LL*(r[j]-j+1)*(j-l[j]+1)); 48 // printf("l[%d] = %d r[%d] = %d cnt = %I64d b[%d] = %d\n",j,l[j],j,r[j],cnt,j,b[j]); 49 ans += cnt*b[j]; 50 } 51 printf("%I64d\n",ans); 52 } 53 } 54 55 int main(){ 56 while(scanf("%d %d",&n,&q) != EOF){ 57 for(int i = 1;i <= n;i++){ 58 scanf("%d",&a[i]); 59 } 60 solve(); 61 } 62 return 0; 63 }
cf 584e 584E - Anton and Ira
重新做以前的题，还是不会捉了----sad---
交换的方法就是 : 如果当前 的 p[i] 和 s[i] 不同，那么就在 p[] 里面找到 p[ed] = s[i]
然后，在 p[i] 到 p[ed] 中，如果存在 p[x],这个p[x] 在 s 中的位置 大于 ed，就交换 ed 和 x
直到换到 ed = i
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 typedef long long LL; 9 typedef pair<int,int> pii; 10 const int maxn = 1e5+5; 11 int p[maxn],s[maxn]; 12 int n; 13 int pos[maxn]; 14 15 void solve(){ 16 vector<pii> ans; 17 LL res = 0; 18 int st = 0,ed = 0; 19 for(int i = 1;i <= n;i++){ 20 ed = 0; 21 if(p[i] == s[i]) continue; 22 for(int j = 1;j <= n;j++){ 23 if(p[j] == s[i]){ 24 ed = j; 25 break; 26 } 27 } 28 // printf("---i = %d ed = %d\n",i,ed); 29 while(ed != i){ 30 for(int j = i;j < ed;j++){ 31 if(pos[p[j]] >= ed){ 32 swap(ed,j); 33 swap(p[ed],p[j]); 34 // printf("=== ed = %d j = %d\n",ed,j); 35 res += 1LL*abs(ed-j); 36 ans.push_back(make_pair(ed,j)); 37 break; 38 } 39 } 40 } 41 } 42 printf("%I64d\n",res); 43 printf("%d\n",ans.size()); 44 for(int i = 0;i < ans.size();i++){ 45 printf("%d %d\n",ans[i].first,ans[i].second); 46 } 47 } 48 49 int main(){ 50 while(scanf("%d",&n) != EOF){ 51 for(int i = 1;i <= n;i++) scanf("%d",&p[i]); 52 for(int i = 1;i <= n;i++) { 53 scanf("%d",&s[i]); 54 pos[s[i]] = i; 55 } 56 solve(); 57 } 58 return 0; 59 }
cf 584b 584B - Kolya and Tanya
当时做这场比赛的时候，，，一看到推公式，，，就跑到oeis 搜---
什么都没搜到----gg---
稍微推一下，就是 3^3n - 7^n
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 typedef long long LL; 8 const int mod = 1e9+7; 9 int n; 10 11 LL Q_pow(LL x,LL y){ 12 LL res = 1; 13 x %= mod; 14 while(y){ 15 if(y&1) res = res*x%mod; 16 x = x*x%mod; 17 y >>= 1; 18 } 19 return res; 20 } 21 22 void solve(){ 23 LL ans = (Q_pow(3,3*n) - Q_pow(7,n)+mod)%mod; 24 printf("%I64d\n",ans); 25 } 26 27 int main(){ 28 while(scanf("%d",&n) != EOF){ 29 solve(); 30 } 31 return 0; 32 }
1.20
寒假起床最晚的一次 T_T
补---题---
cf 496b 496B - Secret Combination
暴力枚举
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 1e5+5; 9 int n; 10 char s[maxn],t[maxn]; 11 char p[10][maxn]; 12 13 void solve(){ 14 vector<string> c; 15 16 for(int i = 0;i <= 9;i++){ 17 for(int j = 1;j <= n;j++){ 18 p[i][j] = ((s[j]-'0')+i)%10+'0'; 19 } 20 } 21 22 /* for(int i = 0;i <= 9;i++){ 23 printf("i = %d p = %s\n",i,p[i]+1); 24 }*/ 25 26 27 for(int i = 0;i <= 9;i++){ 28 for(int k = 0;k <= n-1;k++){ 29 int cnt = 0; 30 for(int l = k+1;l <= n;l++){ 31 t[++cnt] = p[i][l]; 32 } 33 for(int l = 1;l <= k;l++){ 34 t[++cnt] = p[i][l]; 35 } 36 // printf("i = %d k = %d t = %s\n",i,k,t+1); 37 c.push_back(t+1); 38 } 39 } 40 sort(c.begin(),c.end()); 41 printf("%s\n",c[0].c_str()); 42 } 43 44 int main(){ 45 while(scanf("%d",&n) != EOF){ 46 scanf("%s",s+1); 47 solve(); 48 } 49 return 0; 50 }
cf 496c 496C - Removing Columns
....燃烧生命wa水题....sad---
.....wa 了好久....好久....好久阿
给出 n*m 的字母矩阵，为了使得从上到下是字典序递增，至少需要删掉多少列（不是严格递增）
直接判断每一列是不是必须删掉
如果 这一列 是严格上升的，那就不用管
如果这一列出现了s[i][j] < s[i][j-1] ，就必须删掉
如果这一列是非严格上升的，就用 vector 把相同的[st,ed] 存起来，下次就只需要再判断vector 里面的了---
一直 wa 是因为，当判断到这一列必须删掉的时候，，，还把这一列后面的[st,ed] 丢进了 vector里面
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 typedef pair<int,int> pii; 9 char s[105][105]; 10 int n,m; 11 int lie; 12 13 int check(int st,int ed){ 14 int flag = 0; 15 for(int j = st;j <= ed;j++){ 16 if(j == st) continue; 17 if(s[j][lie] < s[j-1][lie]) return 0; 18 if(s[j][lie] == s[j-1][lie]) flag = 1; 19 20 } 21 if(flag == 1) return 2; 22 return 1; 23 } 24 25 void solve(){ 26 if(n == 1){ 27 puts("0"); 28 return; 29 } 30 int res = 0; 31 vector<pii> c[3]; 32 lie = 1; 33 int l = 1,r = n; 34 c[0].push_back(make_pair(l,r)); 35 int key = 0; 36 int tot = 0; 37 while(1){ 38 int cnt = 0; 39 int lb = 0,ub = 0; 40 for(int i = 0;i < c[key].size();i++){ 41 int x = c[key][i].first; 42 int y = c[key][i].second; 43 // printf("i = %d res = %d lie = %d x = %d y = %d\n",i,res,lie,x,y); 44 if(check(x,y) == 1) cnt++; 45 if(check(x,y) == 0){ 46 lb = 1; 47 } 48 if(check(x,y) == 2){ 49 ub = 1; 50 } 51 } 52 //printf("---cnt = %d\n",cnt); 53 if(cnt == c[key].size()) break; 54 if(lb){ 55 res++; 56 c[1-key] = c[key]; 57 } 58 else{ 59 for(int i = 0;i < c[key].size();i++){ 60 int x = c[key][i].first; 61 int y = c[key][i].second; 62 // printf("i = %d res = %d lie = %d x = %d y = %d\n",i,res,lie,x,y); 63 if(check(x,y) == 1) continue; 64 if(check(x,y) == 2){ 65 for(int p = x;p <= y;){ 66 int q = p; 67 while(q<=y && s[q][lie] == s[p][lie])q++; 68 if(q-p > 1){ 69 c[1-key].push_back(make_pair(p,q-1)); 70 } 71 p = q; 72 } 73 } 74 } 75 } 76 c[key].clear(); 77 key = !key; 78 lie++; 79 if(lie == m+1) break; 80 } 81 printf("%d\n",res); 82 } 83 84 int main(){ 85 while(scanf("%d %d",&n,&m) != EOF){ 86 for(int i = 1;i <= n;i++){ 87 scanf("%s",s[i]+1); 88 } 89 solve(); 90 } 91 return 0; 92 }
搜了下题解，，发现自己的做法sb爆了阿
可以用一个数组c[]标记已经不需要再管的行，然后每次碰到一列如果存在下降的情况，就删掉
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 int n,m; 9 char s[105][105]; 10 11 void solve(){ 12 int c[105]; 13 memset(c,0,sizeof(c)); 14 int ans = 0; 15 16 if(n == 1){ 17 puts("0"); 18 return; 19 } 20 for(int i = 1;i <= m;i++){ 21 int ok = 0; 22 for(int j = 2;j <= n;j++){ 23 if(c[j]) continue; 24 if(s[j][i] < s[j-1][i]){ 25 ok = 1; 26 break; 27 } 28 } 29 if(ok){ 30 ans++; 31 } 32 else{ 33 for(int j=2;j <= n;j++){ 34 if(s[j][i] > s[j-1][i]){ 35 c[j] = 1; 36 } 37 } 38 } 39 } 40 printf("%d\n",ans); 41 } 42 43 int main(){ 44 while(scanf("%d %d",&n,&m) != EOF){ 45 for(int i = 1;i <= n;i++){ 46 scanf("%s",s[i]+1); 47 } 48 solve(); 49 } 50 return 0; 51 }
1.21
cf 577b Modulo Sum
不会做的题就再补一次叭--
n > m 的时候，由鸽巢原理
sum[i] 为前 i 项的和对 m 取模
有n 个sum[i] 需要放到 m 个盒子里面，所以肯定有一个盒子里面放了2个sum[i]
所以一定存在 sum[l] = sum[r]
在区间 [l+1,r] 就满足了
然后 n <= m 的时候就dp一下
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<set> 6 using namespace std; 7 8 const int maxn = 1e6+5; 9 int dp[2][maxn],n,m; 10 int a[maxn]; 11 int sum[maxn]; 12 13 void solve(){ 14 memset(dp,0,sizeof(dp)); 15 if(n <= m){ 16 int key = 0; 17 for(int i = 1;i <= n;i++){ 18 dp[key][a[i]%m] = 1; 19 for(int j = 0;j < m;j++){ 20 if(dp[1-key][j]){ 21 dp[key][(j+a[i])%m] = 1; 22 } 23 } 24 for(int j = 0;j < m;j++) dp[1-key][j] = dp[key][j]; 25 key = !key; 26 } 27 if(dp[key][0]) puts("YES"); 28 else puts("NO"); 29 return; 30 } 31 memset(sum,0,sizeof(sum)); 32 for(int i = 1;i <= n;i++){ 33 sum[i] = (sum[i-1]+a[i])%m; 34 } 35 set<int> s; 36 for(int i = 1;i <= n;i++){ 37 if(s.count(sum[i]) == 0){ 38 s.insert(sum[i]); 39 } 40 else{ 41 puts("YES"); 42 return; 43 } 44 } 45 puts("NO"); 46 } 47 48 int main(){ 49 while(scanf("%d %d",&n,&m) != EOF){ 50 for(int i = 1;i <= n;i++){ 51 scanf("%d",&a[i]); 52 } 53 solve(); 54 } 55 return 0; 56 }
然后搜鸽巢原理来看的时候，水一道题
hdu 1205
给出 n 种 物品，个数分别为 a[i],要将每种物品错开来放，问是否可能
最大的a[i] 共有 a[i] - 1个空隙，只需要看剩下的数够不够填这些空隙
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 typedef long long LL; 8 const int maxn = 1e6+5; 9 int a[maxn]; 10 LL sum; 11 int n; 12 13 void solve(){ 14 int maxx = -1; 15 sum = 0; 16 for(int i = 1;i <= n;i++){ 17 maxx = max(maxx,a[i]); 18 sum += 1LL*a[i]; 19 } 20 if(sum-maxx >= maxx-1) puts("Yes"); 21 else puts("No"); 22 } 23 24 int main(){ 25 int T; 26 scanf("%d",&T); 27 while(T--){ 28 scanf("%d",&n); 29 for(int i = 1;i <= n;i++){ 30 scanf("%d",&a[i]); 31 } 32 solve(); 33 } 34 return 0; 35 }
cf 496d 496D - Tennis Game
按照时间给出 n 小局比赛的胜负，1 为A赢，2为 B赢，每赢一局得1分
如果一个人赢够了 t 分，那么这一大局比赛结束，如果一个人赢够了 s 大局，整个比赛就结束
求所有合法的s,t
-----先自己想的是枚举t，二分 s，可是不懂判断 对应一个 t ，这个s是不是合法---
----然后可以这样
先预处理出sum1[i] 表示前 i 项有多少个1
sum2[i]表示前 i 项有多少个2
枚举t,然后直接去计算 s
注意的是，如果一个人这场比赛赢的话，那么最后一小局也是他赢
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 typedef pair<int,int> pii; 9 const int maxn = 1e5+5; 10 int sum1[maxn],sum2[maxn]; 11 int a[maxn],n; 12 13 int work(int t){ 14 int pos = 0,p1,p2,s1=0,s2=0,win; 15 while(pos <= n){ 16 p1 = lower_bound(sum1+1,sum1+n+1,sum1[pos]+t) - sum1; 17 p2 = lower_bound(sum2+1,sum2+n+1,sum2[pos]+t) - sum2; 18 // printf("t = %d p1 = %d p2 = %d s1 = %d s2 = %d\n",t,p1,p2,s1,s2); 19 if(p1 < p2){ 20 s1++; 21 win = 1; 22 } 23 pos = min(p1,p2); 24 if(p1 > p2){ 25 s2++; 26 win = 2; 27 } 28 if(pos == n) break; 29 if(pos > n) return 0; 30 } 31 // printf("-- t= %d win = %d s1 = %d s2 = %d pos = %d\n",t,win,s1,s2,pos); 32 if((win == 1 && s1 > s2 ) || (win == 2 && s1 < s2)) return max(s1,s2); 33 return 0; 34 } 35 36 void solve(){ 37 memset(sum1,0,sizeof(sum1)); 38 memset(sum2,0,sizeof(sum2)); 39 for(int i = 1;i <= n;i++){ 40 if(a[i] == 1) sum1[i] = sum1[i-1]+1; 41 else sum1[i] = sum1[i-1]; 42 if(a[i] == 2) sum2[i] = sum2[i-1]+1; 43 else sum2[i] = sum2[i-1]; 44 } 45 46 /*for(int i = 1;i <= n;i++){ 47 printf("sum1[%d] = %d sum2[%d] = %d\n",i,sum1[i],i,sum2[i]); 48 }*/ 49 50 vector<pii> ans; 51 for(int i = 1;i <= n;i++){ 52 int s = work(i); 53 if(s == 0) continue; 54 ans.push_back(make_pair(s,i)); 55 } 56 sort(ans.begin(),ans.end()); 57 printf("%d\n",ans.size()); 58 for(int i = 0;i < ans.size();i++){ 59 printf("%d %d\n",ans[i].first,ans[i].second); 60 } 61 } 62 63 int main(){ 64 while(scanf("%d",&n) != EOF){ 65 for(int i = 1;i <= n;i++){ 66 scanf("%d",&a[i]); 67 } 68 solve(); 69 } 70 return 0; 71 }
hdu 1160
找出最长 的满足
w[i] < w[i+1] < ---- < w[x]
s[i] > s[i+1] > --- > s[x]
输出长度和路径
想到暑假做的连通分量的缩点之后求DAG上的最长路，
按照关系建好图，dfs一下，找出最大值，再根据这个最大值去找路径
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 const int maxn = 1e3+5; 9 10 struct node{ 11 int x,y; 12 int id; 13 }p[maxn]; 14 15 vector<int> g[maxn]; 16 int n; 17 int dp[maxn]; 18 19 int dfs(int u){ 20 if(dp[u]) return dp[u]; 21 for(int i = 0;i < g[u].size();i++){ 22 int v = g[u][i]; 23 dp[u] = max(dp[u],dfs(v)); 24 } 25 return dp[u] = dp[u]+1; 26 } 27 28 void solve(){ 29 for(int i = 1;i <= n;i++) g[i].clear(); 30 memset(dp,0,sizeof(dp)); 31 /* for(int i = 1;i <= n;i++){ 32 printf("p[%d].x = %d p[%d].y = %d\n",i,p[i].x,i,p[i].y); 33 }*/ 34 35 for(int i = 1;i <= n;i++){ 36 for(int j = 1;j <= n;j++){ 37 if(i == j) continue; 38 if(p[i].x < p[j].x && p[i].y > p[j].y){ 39 g[i].push_back(j); 40 } 41 } 42 } 43 44 /* for(int i = 1;i <= n;i++){ 45 printf("---i = %d ",i); 46 for(int j = 0;j < g[i].size();j++){ 47 printf("%d ",g[i][j]); 48 } 49 printf("\n"); 50 }*/ 51 52 int res = -1; 53 for(int i = 1;i <= n;i++){ 54 int tmp = dfs(i); 55 res = max(res,tmp); 56 // printf("i = %d tmp = %d\n",i,tmp); 57 } 58 59 // for(int i = 1;i <= n;i++){ 60 // printf("--dp[%d] = %d\n",i,dp[i]); 61 //} 62 63 vector<int> l; 64 int st = 0,maxx = -1; 65 for(int i = 1;i <= n;i++){ 66 if(dp[i] > maxx){ 67 st = i; 68 maxx = dp[i]; 69 } 70 } 71 int r = res; 72 l.push_back(st); 73 // printf("--st = %d\n",st); 74 while(r>=1){ 75 for(int i = 0;i < g[st].size();i++){ 76 int v = g[st][i]; 77 if(dp[v] == r){ 78 st = v; 79 // printf("v = %d\n",v); 80 l.push_back(v); 81 break; 82 } 83 } 84 r--; 85 } 86 printf("%d\n",l.size()); 87 for(int i = 0;i < l.size();i++){ 88 printf("%d\n",l[i]); 89 } 90 } 91 92 int main(){ 93 // freopen("in.txt","r",stdin); 94 //freopen("out.txt","w",stdout); 95 n = 1; 96 while(scanf("%d %d",&p[n].x,&p[n].y) != EOF){ 97 n++; 98 } 99 solve(); 100 return 0; 101 }
cf 495b 495B - Modular Equations
求满足 a %x = b 的x有多少个
a = b 无穷多个
a < b 0个
a > b
可以化简成 a = kx + b
所以 kx = a-b
就在1 到 sqrt(a-b) 之间枚举k,记录答案
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 #include<map> 7 #include<vector> 8 using namespace std; 9 10 int a,b; 11 const int maxn = 1e6+5; 12 13 void solve(){ 14 if(a == b){ 15 puts("infinity"); 16 return; 17 } 18 if(a < b){ 19 puts("0"); 20 return; 21 } 22 int ans = 0; 23 int n=a-b; 24 for(int i = 1;i <= sqrt(n);i++){ 25 if(n%i == 0){ 26 int x = i; 27 int y = n/i; 28 if(x == y && x > b){ 29 ans++; 30 continue; 31 } 32 if(x > b) { 33 ans++; 34 // printf("--x = %d\n",x); 35 } 36 if(y > b) { 37 ans++; 38 // printf("---y = %d\n",y); 39 } 40 } 41 } 42 43 printf("%d\n",ans); 44 } 45 46 int main(){ 47 48 while(scanf("%d %d",&a,&b) != EOF){ 49 solve(); 50 } 51 return 0; 52 }
cf 495c 495C - Treasure
括号匹配，先消去一次后，再补
每次取s.top()的时候，记得判断下size阿，老是re
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 #include<stack> 7 using namespace std; 8 9 const int maxn = 1e5+5; 10 char t[maxn]; 11 int n; 12 int ans[maxn]; 13 14 void solve(){ 15 n = strlen(t+1); 16 stack<char> s; 17 int ub = 0; 18 for(int i = 1;i <= n;i++){ 19 if(t[i] == '#'){ 20 s.push(t[i]); 21 ub++; 22 continue; 23 } 24 if(t[i] == '('){ 25 s.push(t[i]); 26 } 27 if(t[i] == ')'){ 28 if(s.size() != 0){ 29 char c = s.top(); 30 int cnt = 0; 31 while(c == '#'){ 32 s.pop(); 33 cnt++; 34 if(s.size() == 0) break; 35 c = s.top(); 36 } 37 // printf("---cnt = %d\n",cnt); 38 if(c == '(') { 39 s.pop(); 40 for(int j = 1;j <= cnt;j++){ 41 s.push('#'); 42 } 43 } 44 else{ 45 puts("-1"); 46 return; 47 } 48 } 49 else{ 50 puts("-1"); 51 return; 52 } 53 } 54 } 55 vector<char> cp; 56 while(!s.empty()){ 57 cp.push_back(s.top()); 58 s.pop(); 59 } 60 reverse(cp.begin(),cp.end()); 61 /* for(int i = 0;i < cp.size();i++){ 62 printf("%c",cp[i]); 63 } 64 printf("\n");*/ 65 66 int lb = 0; 67 memset(ans,0,sizeof(ans)); 68 for(int i = 0;i < cp.size();i++){ 69 if(cp[i] == '('){ 70 s.push(cp[i]); 71 } 72 if(cp[i] == '#'){ 73 if(s.size() == 0){ 74 puts("-1"); 75 return; 76 } 77 lb++; 78 if(lb != ub){ 79 ans[lb] = 1; 80 s.pop(); 81 } 82 else{ 83 ans[lb] = s.size(); 84 while(!s.empty()) s.pop(); 85 } 86 } 87 if(cp[i] == ')'){ 88 puts("-1"); 89 return; 90 } 91 } 92 if(!s.empty()){ 93 puts("-1"); 94 return; 95 } 96 for(int i = 1;i <= lb;i++){ 97 printf("%d\n",ans[i]); 98 } 99 } 100 101 int main(){ 102 while(scanf("%s",t+1) != EOF){ 103 solve(); 104 } 105 return 0; 106 }
1.22
cf 620c C - Pearls in a Row
contain 是包含的意思，至少包含有2个不同的数
第一次wa 是因为 没有考虑到最长--
然后后几次 wa 是因为 contain
sad-------------------------------------
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 #include<set> 7 using namespace std; 8 9 typedef pair<int,int> pii; 10 typedef long long LL; 11 const int maxn = 5e5+5; 12 int n,m; 13 int a[maxn]; 14 15 void solve(){ 16 vector<pii> ans; 17 int l = 1,r = 1; 18 for(l = 1;l <= n;){ 19 r = l; 20 set<int> s; 21 while(r <= n && s.count(a[r]) == 0){ 22 s.insert(a[r]); 23 r++; 24 } 25 if(r == n+1) break; 26 ans.push_back(make_pair(l,r)); 27 l = r+1; 28 } 29 if(ans.size() == 0){ 30 puts("-1"); 31 return; 32 } 33 34 int sz = ans.size()-1; 35 ans[sz].second = n; 36 37 printf("%d\n",ans.size()); 38 for(int i = 0;i < ans.size();i++){ 39 printf("%d %d\n",ans[i].first,ans[i].second); 40 } 41 } 42 43 int main(){ 44 while(scanf("%d",&n) != EOF){ 45 for(int i = 1;i <= n;i++){ 46 scanf("%d",&a[i]); 47 } 48 solve(); 49 } 50 return 0; 51 }
cf 620d D - Professor GukiZ and Two Arrays
k = 0,abs (sa - sb)
k = 1, n*m 枚举
k = 2 的时候，用 map<LL,pair<int,int> > 把每一对的 a[i] ,a[j] 存起来
枚举b[i] ,b[j],去找最接近 sa-sb + 2*(b[i]+b[j]) 的 2*（a[i]+a[j]） 来更新答案
----今早发现 wa 了--
应该是找到 it,和 it-- 叭
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 #include<vector> 7 #include<map> 8 using namespace std; 9 10 typedef long long LL; 11 typedef pair<int,int> pii; 12 const int maxn = 5e3+5; 13 int n,m; 14 LL a[maxn],b[maxn]; 15 16 struct node{ 17 int num; 18 LL sum; 19 int l,r; 20 int ll,rr; 21 }p[5]; 22 23 void solve(){ 24 LL res = 1e13+7; 25 LL sa = 0,sb = 0; 26 for(int i = 1;i <= n;i++){ 27 sa += a[i]; 28 } 29 for(int i = 1;i <= m;i++){ 30 sb += b[i]; 31 } 32 p[1].num = 0; 33 p[1].sum = abs(sa-sb); 34 p[2].num = 1; 35 p[3].num = 2; 36 for(int i = 1;i <= n;i++){ 37 for(int j = 1;j <= m;j++){ 38 LL tmp = abs(sa-sb+2*b[j]-2*a[i]); 39 // printf("i = %d j = %d tmp = %I64d\n",i,j,tmp); 40 if(tmp < res){ 41 res = tmp; 42 p[2].sum = res; 43 p[2].l = i;p[2].r = j; 44 } 45 } 46 } 47 int num = 2; 48 if(n >= 2 && m >= 2){ 49 map<LL,pair<int,int> > h; 50 for(int i = 1;i <= n;i++){ 51 for(int j = i+1;j <= n;j++){ 52 h[2*(a[i]+a[j])] = make_pair(i,j); 53 } 54 } 55 res = 1e13+7; 56 map<LL,pair<int,int> >::iterator it; 57 58 for(int i = 1;i <= m;i++){ 59 for(int j = i+1;j <= m;j++){ 60 LL tmp = sa-sb+2*(b[i]+b[j]); 61 it = h.lower_bound(tmp); 62 for(int k = 0;k < 2;k++){ 63 if(it == h.end()){ 64 it--; 65 continue; 66 } 67 if(abs(tmp-it->first) < res){ 68 res = abs(tmp-it->first); 69 p[3].sum = res; 70 p[3].l = it->second.first;p[3].r = i; 71 p[3].ll = it->second.second;p[3].rr = j; 72 } 73 if(it == h.begin()) break; 74 it--; 75 } 76 } 77 num = 3; 78 } 79 } 80 81 res = 1e13+7; 82 for(int i = 1;i <= num;i++){ 83 res = min(res,p[i].sum); 84 } 85 // printf("--res = %I64d\n",res); 86 for(int i = 1;i <= num;i++){ 87 if(p[i].sum == res){ 88 printf("%I64d\n",p[i].sum); 89 if(i == 1){ 90 puts("0"); 91 return; 92 } 93 if(i == 2){ 94 puts("1"); 95 printf("%d %d\n",p[i].l,p[i].r); 96 return; 97 } 98 puts("2"); 99 printf("%d %d\n",p[i].l,p[i].r); 100 printf("%d %d\n",p[i].ll,p[i].rr); 101 } 102 } 103 104 } 105 106 int main(){ 107 while(scanf("%d",&n) != EOF){ 108 for(int i = 1;i <= n;i++){ 109 scanf("%I64d",&a[i]); 110 } 111 scanf("%d",&m); 112 for(int i = 1;i <= m;i++){ 113 scanf("%I64d",&b[i]); 114 } 115 solve(); 116 } 117 return 0; 118 }
1.23
补题
hdu 5612 Baby Ming and Matrix games
知道搜一下就好了，可是一直wa，果然不会搜索--撒都不会----
先是用的double
看bc群里说的会卡double
然后，，搜索真的不会--
搜之前
vis[xx][yy] = 1;
dfs(xx,yy);
搜完再把它改回来阿-唉
然后，就分别用分子，分母来加减乘除
最后，还wa了，是因为 ok 没有初始化
---sad----
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<set> 6 using namespace std; 7 8 typedef long long LL; 9 char g[105][105]; 10 int vis[105][105]; 11 int n,m; 12 LL sum; 13 int ok; 14 int dx[4]={1,-1,0,0}; 15 int dy[4]={0,0,1,-1}; 16 17 LL gcd(LL a,LL b){ 18 return (!b)? a:gcd(b,a%b); 19 } 20 21 void dfs(int x,int y,LL fz,LL fm){ 22 if(fz == sum && fm == 1){ 23 ok = 1; 24 return; 25 } 26 for(int i = 0;i < 4;i++){ 27 int xx = x+2*dx[i]; 28 int yy = y+2*dy[i]; 29 int nx = x+dx[i]; 30 int ny = y+dy[i]; 31 if(xx<1||xx>n||yy<1||yy>m||vis[xx][yy]) continue; 32 LL nfz = fz,nfm = fm; 33 if(g[nx][ny] == '+'){ 34 nfz = nfz+(g[xx][yy]-'0')*fm; 35 } 36 if(g[nx][ny] == '-'){ 37 nfz = nfz-(g[xx][yy]-'0')*fm; 38 } 39 if(g[nx][ny] == '*'){ 40 nfz = nfz*(g[xx][yy]-'0'); 41 } 42 if(g[nx][ny] == '/'){ 43 if(g[xx][yy] == '0') continue; 44 nfm = nfm*(g[xx][yy]-'0'); 45 } 46 vis[xx][yy] = 1; 47 LL gc = gcd(nfz,nfm); 48 nfz = nfz/gc;nfm = nfm/gc; 49 dfs(xx,yy,nfz,nfm); 50 vis[xx][yy] = 0; 51 } 52 } 53 54 void solve(){ 55 LL fz,fm; 56 ok = 0; 57 for(int i = 1;i <= n;i++){ 58 for(int j = 1;j <= m;j++){ 59 if(i%2 && j%2){ 60 memset(vis,0,sizeof(vis)); 61 vis[i][j] = 1; 62 fz = g[i][j]-'0'; 63 fm = 1; 64 dfs(i,j,fz,fm); 65 if(ok == 1){ 66 puts("Possible"); 67 return; 68 } 69 vis[i][j] = 0; 70 } 71 } 72 } 73 puts("Impossible"); 74 } 75 76 int main(){ 77 int T; 78 scanf("%d",&T); 79 while(T--){ 80 scanf("%d %d %I64d",&n,&m,&sum); 81 for(int i = 1;i <= n;i++){ 82 scanf("%s",g[i]+1); 83 } 84 solve(); 85 } 86 return 0; 87 }
1.24
补题
cf 617c Watering Flowers
就过了6个样例还给过pretest，诶
犯了两个错，去扫最大值得时候，把rr 初始化为 -1，因为更新答案的时候是用 rr+RR，在有些时候，就会让答案少1
然后，只枚举了一个 R,算另一个r
应该两个圆的半径都分别去枚举的时候，取最优的
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 typedef long long LL; 9 const int maxn = 1e5+5; 10 int a[maxn]; 11 int n,m; 12 13 struct node{ 14 LL r,R; 15 LL x,y; 16 }p[maxn]; 17 18 LL x1,x2,y1,y2; 19 20 void solve(){ 21 LL ans = 1e18+7; 22 for(int i = 1;i <= n;i++){ 23 ans = max(ans,p[i].r); 24 ans = max(ans,p[i].R); 25 } 26 27 /* for(int i = 1;i <= n;i++){ 28 printf("p[%d].r = %I64d R = %I64d\n",i,p[i].r,p[i].R); 29 }*/ 30 31 for(int i = 1;i <= n;i++){ 32 LL RR = p[i].R; 33 LL rr = 0; 34 // printf("--i == %d RR = %I64d\n",i,RR); 35 for(int j = 1;j <= n;j++){ 36 if(p[j].R <= RR) continue; 37 if(i == j) continue; 38 rr = max(rr,p[j].r); 39 // printf("i = %d j = %d rr = %I64d RR = %I64d\n",i,j,rr,RR); 40 } 41 ans = min(ans,rr+RR); 42 } 43 44 for(int i = 1;i <= n;i++){ 45 LL rr = p[i].r; 46 LL RR = 0; 47 // printf("--i == %d RR = %I64d\n",i,RR); 48 for(int j = 1;j <= n;j++){ 49 if(p[j].r <= rr) continue; 50 if(i == j) continue; 51 RR = max(RR,p[j].R); 52 // printf("i = %d j = %d rr = %I64d RR = %I64d\n",i,j,rr,RR); 53 } 54 ans = min(ans,rr+RR); 55 } 56 57 printf("%I64d\n",ans); 58 } 59 60 int main(){ 61 while(scanf("%d %I64d %I64d %I64d %I64d",&n,&x1,&y1,&x2,&y2) != EOF){ 62 for(int i = 1;i <= n;i++){ 63 scanf("%I64d %I64d",&p[i].x,&p[i].y); 64 p[i].r = (p[i].x-x1)*(p[i].x-x1)+(p[i].y-y1)*(p[i].y-y1); 65 p[i].R = (p[i].x-x2)*(p[i].x-x2)+(p[i].y-y2)*(p[i].y-y2); 66 } 67 solve(); 68 } 69 return 0; 70 }
cf 617d D - Polyline
被hack 之后，输出 2 的情况还是没有想完
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 8 typedef long long LL; 9 const int maxn = 1e5+5; 10 int a[maxn]; 11 int n,m; 12 int x1,x2,x3,y1,y2,y3; 13 14 void solve(){ 15 if((x1 == x2 &&x2 == x3)||(y1 == y2 && y2 == y3)){ 16 puts("1"); 17 return; 18 } 19 20 if((y1 == y2)&&(x3>=max(x1,x2) ||x3 <= min(x1,x2))){ 21 puts("2"); 22 return; 23 } 24 25 if((y2 == y3)&&(x1>=max(x2,x3) ||x1 <= min(x2,x3))){ 26 puts("2"); 27 return; 28 } 29 30 if((y1 == y3)&&(x2>=max(x1,x3) ||x2 <= min(x1,x3))){ 31 puts("2"); 32 return; 33 } 34 35 if((x3 == x2)&&(y1>=max(y2,y3) ||y1 <= min(y2,y3))){ 36 puts("2"); 37 return; 38 } 39 40 if((x1 == x3)&&(y2>=max(y1,y3) ||y2 <= min(y1,y3))){ 41 puts("2"); 42 return; 43 } 44 45 if((x1 == x2)&&(y3>=max(y1,y2) ||y3 <= min(y1,y2))){ 46 puts("2"); 47 return; 48 } 49 puts("3"); 50 } 51 52 int main(){ 53 while(scanf("%d %d",&x1,&y1) != EOF){ 54 scanf("%d %d",&x2,&y2); 55 scanf("%d %d",&x3,&y3); 56 solve(); 57 } 58 return 0; 59 }
hdu 5145 NPY and girls
给出 n 个数，m 个询问 [l,r] 里面全排列的个数
照着这篇敲的
http://blog.csdn.net/bossup/article/details/39236275
其实还是不会用阿
然后 re 是因为 最后再乘以分母的逆元，会超过打表求出来的Inv数组，需要重新算一下
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<vector> 7 using namespace std; 8 9 typedef long long LL; 10 const int maxn = 1e5+5; 11 const int mod = 1e9+7; 12 int n,m; 13 LL num[maxn]; 14 int col[maxn],pos[maxn]; 15 LL afac[maxn+10],fac[maxn+10]; 16 LL ans; 17 LL res[maxn]; 18 19 struct node{ 20 int l,r; 21 int id; 22 }q[maxn]; 23 24 bool cmp(node a,node b){ 25 if(pos[a.l] == pos[b.l]) 26 return a.r<b.r; 27 return pos[a.l] < pos[b.l]; 28 } 29 30 LL Q_pow(LL x,LL y){ 31 LL res = 1; 32 x %= mod; 33 while(y){ 34 if(y&1) res = res*x%mod; 35 x = x*x%mod; 36 y >>= 1; 37 } 38 return res; 39 } 40 41 void Pre(){ 42 fac[0]=1; 43 for(int i = 1;i <= maxn;i++){ 44 fac[i] = fac[i-1]*(LL)i%mod; 45 } 46 afac[maxn]=Q_pow(fac[maxn],mod-2); 47 for(int i = maxn;i>=1;i--){ 48 afac[i-1] = afac[i]*i%mod; 49 } 50 } 51 52 void update(int x,int d){ 53 ans = ans*afac[num[col[x]]]%mod; 54 num[col[x]] += d; 55 ans = ans*fac[num[col[x]]]%mod; 56 } 57 58 long long Inv(long long x)///mod为质数 59 { 60 long long r, y; 61 for(r = 1, y = mod - 2; y; x = x * x % mod, y >>= 1) 62 (y & 1) && (r = r * x % mod); 63 return r; 64 } 65 66 int main(){ 67 Pre(); 68 int bk,pl,pr,id; 69 int T; 70 scanf("%d",&T); 71 while(T--){ 72 scanf("%d %d",&n,&m); 73 memset(num,0,sizeof(num)); 74 bk = ceil(sqrt(1.0*n)); 75 for(int i = 1;i <= n;i++){ 76 scanf("%d",&col[i]); 77 pos[i]=(i-1)/bk; 78 } 79 for(int i = 0;i < m;i++){ 80 scanf("%d %d",&q[i].l,&q[i].r); 81 q[i].id = i; 82 } 83 sort(q,q+m,cmp); 84 pl = 1,pr = 0; 85 ans = 1; 86 memset(res,0,sizeof(res)); 87 for(int i = 0;i < m;i++){ 88 id = q[i].id; 89 if(q[i].l == q[i].r){ 90 res[id] = 1; 91 continue; 92 } 93 if(pr < q[i].r){ 94 for(int j = pr+1;j <= q[i].r;j++){ 95 update(j,1); 96 } 97 } 98 else{ 99 for(int j = pr;j >q[i].r;j--){ 100 update(j,-1); 101 } 102 } 103 pr = q[i].r; 104 if(pl<q[i].l){ 105 for(int j = pl;j<q[i].l;j++){ 106 update(j,-1); 107 } 108 } 109 else{ 110 for(int j=pl-1;j>=q[i].l;j--){ 111 update(j,1); 112 } 113 } 114 pl = q[i].l; 115 LL fz = (q[i].r-q[i].l+1); 116 LL tmp = fac[fz]*Inv(ans)%mod; 117 res[id] = tmp; 118 } 119 for(int i = 0;i < m;i++){ 120 printf("%I64d\n",res[i]); 121 } 122 } 123 return 0; 124 }
hdu 5613Baby Ming and Binary image
看题解补的，暴力枚举最左边一列，2^n-2，然后算这 2^n-2个矩阵里面有多少个满足的
wa了好几次，是因为递推完一个矩阵的时候，要验算一下最后一行是不是都是0
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<vector> 6 7 int n,m; 8 int g[55][105]; 9 int res[55][105]; 10 int a[55][105]; 11 int dx[8]={0,-1,-1,-1,0,1,1,0}; 12 int dy[8]={1,1,0,-1,-1,-1,0,0}; 13 14 int ok(){ 15 /* for(int i = 1;i <= n;i++){ 16 for(int j = 1;j <= m;j++){ 17 printf("%d ",a[i][j]); 18 } 19 printf("\n"); 20 } 21 printf("----newnewnewnwew\n");*/ 22 23 for(int i = 1;i <= n;i++){ 24 for(int j = 1;j <= m;j++){ 25 int tmp = 0; 26 for(int k = 0;k < 8;k++){ 27 int nx = i+dx[k]; 28 int ny = j+dy[k]; 29 if(nx<1||nx>n||ny<1||ny>m) continue; 30 tmp += a[nx][ny]; 31 } 32 // printf("i = %d j = %d tmp=%d\n",i,j,tmp); 33 if(g[i][j] == tmp) a[i+1][j+1]=0; 34 else if(g[i][j]== tmp+1){ 35 if(i+1 <= n && j+1 <= m) a[i+1][j+1]=1; 36 else return 0; 37 } 38 else{ 39 return 0; 40 } 41 /* printf("----debug----\n"); 42 for(int i = 1;i <= n;i++){ 43 for(int j = 1;j <= m;j++){ 44 printf("%d ",a[i][j]); 45 } 46 printf("----\n"); 47 } 48 printf("----debug----\n");*/ 49 } 50 } 51 for(int j=1;j <= m;j++){ 52 if(a[n][j] != 0) return 0; 53 } 54 return 1; 55 } 56 57 void solve(){ 58 int N = n-2; 59 int c = 0; 60 for(int i = 0;i < (1<<N);i++){ 61 memset(a,0,sizeof(a)); 62 for(int j = 0;j < N;j++){ 63 if(i&(1<<j)){ 64 a[j+2][1]=1; 65 } 66 } 67 if(ok()) { 68 c++; 69 memcpy(res,a,sizeof(a)); 70 } 71 } 72 // printf("cccc = %d\n",c); 73 if(c == 1){ 74 for(int i = 1;i <= n;i++){ 75 for(int j = 1;j <= m;j++){ 76 if(j!=m) printf("%d ",res[i][j]); 77 else printf("%d\n",res[i][j]); 78 } 79 } 80 } 81 else if(c == 0){ 82 puts("Impossible"); 83 } 84 else{ 85 puts("Multiple"); 86 } 87 } 88 89 int main(){ 90 int T; 91 92 scanf("%d",&T); 93 while(T--){ 94 scanf("%d %d",&n,&m); 95 for(int i = 1;i <= n;i++){ 96 for(int j = 1;j <= m;j++){ 97 scanf("%d",&g[i][j]); 98 } 99 } 100 solve(); 101 } 102 return 0; 103 }