寒假第一周 1.11 --- 1.17
比暑假还长的寒假---
1.11
cf 608d
http://codeforces.com/contest/608/problem/D
不会做，看的题解
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 505; 8 int a[maxn],dp[maxn][maxn]; 9 int n; 10 const int INF = (1<<30)-1; 11 12 int dfs(int l,int r){ 13 if(l >= r) return 1; 14 if(dp[l][r]) return dp[l][r]; 15 int res = INF; 16 if(a[l] == a[r]) res = min(res,dfs(l+1,r-1)); 17 for(int i = l;i < r;i++){ 18 res = min(res,dfs(l,i) + dfs(i+1,r)); 19 } 20 return dp[l][r] = res; 21 } 22 23 int main(){ 24 while(scanf("%d",&n) != EOF){ 25 for(int i = 1;i <= n;i++) scanf("%d",&a[i]); 26 memset(dp,0,sizeof(dp)); 27 printf("%d\n",dfs(1,n)); 28 } 29 return 0; 30 }
cf 296c
http://codeforces.com/problemset/problem/296/C
先以为得用 线段树的区间更新，，后来发现可以不用--
把区间更新学一下下叭---再来写一下
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 #define lp (p<<1) 7 #define rp (p << 1|1) 8 #define getmid(l,r) (l+(r-l)/2) 9 10 typedef long long LL; 11 const int maxn = 5e5+5; 12 13 struct node{ 14 int l,r; 15 int d; 16 }b[maxn]; 17 18 node c[maxn]; 19 int cnt[maxn]; 20 int p[maxn]; 21 LL a[maxn]; 22 LL add[maxn]; 23 LL q[maxn]; 24 int n,m,k; 25 26 void solve(){ 27 memset(cnt,0,sizeof(cnt)); 28 for(int i = 1;i <= k;i++){ 29 cnt[c[i].l]++; 30 cnt[c[i].r+1]--; 31 } 32 33 /*for(int i = 1;i <= m;i++){ 34 printf("cnt[%d] = %d\n",i,cnt[i]); 35 }*/ 36 37 for(int i = 1;i <= m;i++) cnt[i] = cnt[i-1] + cnt[i]; 38 39 /*for(int i = 1;i <= m;i++){ 40 printf("cnt[%d] = %d\n",i,cnt[i]); 41 }*/ 42 memset(add,0,sizeof(add)); 43 for(int i = 1;i <= m;i++){ 44 if(cnt[i] == 0) continue; 45 add[b[i].l] += 1LL*cnt[i]*b[i].d; 46 add[b[i].r+1] -= 1LL*cnt[i]*b[i].d; 47 } 48 49 for(int i = 1;i <= n;i++) add[i] = add[i-1] + add[i]; 50 for(int i = 1;i <= n;i++){ 51 a[i] += add[i]; 52 } 53 for(int i = 1;i <= n;i++) printf("%I64d ",a[i]); 54 printf("\n"); 55 } 56 57 int main(){ 58 while(scanf("%d %d %d",&n,&m,&k) != EOF){ 59 for(int i = 1;i <= n;i++) scanf("%I64d",&a[i]); 60 for(int i = 1;i <= m;i++){ 61 scanf("%d %d %d",&b[i].l,&b[i].r,&b[i].d); 62 } 63 for(int i = 1;i <= k;i++){ 64 scanf("%d %d",&c[i].l,&c[i].r); 65 } 66 solve(); 67 } 68 return 0; 69 }
1.12
cf 616d
http://codeforces.com/contest/616/problem/D
发现真的不会写尺取阿---
wa了一下午 ---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 const int maxn = 1e6+5; 11 int a[maxn]; 12 int n,k; 13 14 void solve(){ 15 pii res; 16 multiset<int> ss; 17 set<int> s; 18 int len = 0; 19 int j = 1,i = 1; 20 while(j <= n){ 21 while(j <= n && s.size() <= k){ 22 if(s.size() == k){ 23 if(!s.count(a[j])) break; 24 } 25 s.insert(a[j]); 26 // printf("i = %d j = %d s.size() = %d\n",i,j,s.size()); 27 ss.insert(a[j]); 28 if(s.size() > k) break; 29 j++; 30 31 } 32 if(j-i > len && s.size() <= k){ 33 len = j-i; 34 res = make_pair(i,j-1); 35 } 36 if(ss.count(a[i]) == 1){ 37 //printf("------\n"); 38 s.erase(a[i]); 39 } 40 int x = a[i]; 41 multiset<int>::iterator pos = ss.find(x); 42 ss.erase(pos); 43 i++; 44 // printf("=== i = %d j = %d\n",i,j); 45 } 46 printf("%d %d\n",res.first,res.second); 47 } 48 49 int main(){ 50 while(scanf("%d %d",&n,&k) != EOF){ 51 for(int i = 1;i <= n;i++) scanf("%d",&a[i]); 52 solve(); 53 } 54 return 0; 55 }
1.13
cf 615c
http://codeforces.com/problemset/problem/615/C
还是不懂dp的做法，，最长公共前缀是什么阿--sad--
每次贪心取能够匹配到的最长的子串，一直到匹配完b串
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 = 3005; 10 char a[maxn],b[maxn],ra[maxn]; 11 int lena,lenb; 12 int stl,edl; 13 int str,edr; 14 int pos; 15 16 void worklb(char *s){ 17 int res = 0; 18 stl = 0,edl = 0; 19 int len = 0; 20 for(int i = 1;i <= lena;i++){ 21 for(int j = i;j <= lena;j++){ 22 if(s[j] != b[pos+j-i]){ 23 break; 24 } 25 if(j-i+1 > len){ 26 stl = i; 27 edl = j; 28 len = j-i+1; 29 } 30 //printf("- i = %d j = %d\n",i,j); 31 } 32 } 33 } 34 35 void workub(char *s){ 36 int res = 0; 37 str = 0,edr = 0; 38 int len = 0; 39 for(int i = 1;i <= lena;i++){ 40 for(int j = i;j <= lena;j++){ 41 if(s[j] != b[pos+j-i]){ 42 break; 43 } 44 if(j-i+1 > len){ 45 len = j-i+1; 46 str = i; 47 edr = j; 48 } 49 // printf("-- i = %d j = %d\n",i,j); 50 } 51 } 52 } 53 54 void solve(){ 55 vector<pii> ans; 56 for(int i = 1;i <= lena;i++) ra[i] = a[lena-i+1]; 57 pos = 1; 58 int lb,ub; 59 int cnt = 0; 60 while(pos <= lenb){ 61 worklb(a); 62 workub(ra); 63 //printf("stl = %d edl = %d\n",stl,edl); 64 //printf("str = %d edr = %d\n",str,edr); 65 if(edl == 0 && stl == 0 && str == 0 && edr == 0){ 66 puts("-1"); 67 return; 68 } 69 if(edl-stl >= edr-str){ 70 ans.push_back(make_pair(stl,edl)); 71 pos += edl-stl+1; 72 // printf("-pos = %d\n",pos); 73 } 74 else{ 75 ans.push_back(make_pair(lena-str+1,lena-edr+1)); 76 pos += edr-str+1; 77 // printf("--pos = %d\n",pos); 78 } 79 // printf("pos = %d\n",pos); 80 } 81 printf("%d\n",ans.size()); 82 for(int i = 0;i < ans.size();i++){ 83 printf("%d %d\n",ans[i].first,ans[i].second); 84 } 85 } 86 87 int main(){ 88 scanf("%s",a+1); 89 scanf("%s",b+1); 90 lena = strlen(a+1); 91 lenb = strlen(b+1); 92 solve(); 93 }
cf 611e
http://codeforces.com/contest/611/problem/E
先是将每场战争从大到小排序，然后分别从两边贪
wa 8 ---------
然后，，，能力值最弱的优先，耗费的人数最少的优先
然后，在打败最强的之后，剩下的枪手的分配，应该是找到能力值和它最相近的t[i]
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<set> 6 using namespace std; 7 8 const int maxn = 2e5+5; 9 int t[maxn]; 10 int n,a,b,c; 11 12 void solve(){ 13 int d[3]; 14 d[0] = a;d[1] = b;d[2] = c; 15 sort(d,d+3); 16 a = d[0];b = d[1];c = d[2]; 17 sort(t+1,t+n+1); 18 19 /* for(int i = 1;i <= n;i++) printf("%d ",t[i]); 20 printf("\n");*/ 21 multiset<int> s; 22 for(int i = 1;i <= n;i++){ 23 s.insert(t[i]); 24 } 25 if((a+b+c) < t[n]){ 26 puts("-1"); 27 return; 28 } 29 int cc = 0; 30 multiset<int>::iterator it; 31 multiset<int>::iterator itt; 32 while(s.size() != 0){ 33 it = s.end();it--; 34 int x = *it; 35 int y; 36 s.erase(it); 37 cc++; 38 if(s.size() == 0) break; 39 if(a >= x){ 40 if(s.size() != 0) { 41 itt = s.begin(); 42 s.erase(itt); 43 } 44 if(s.size() != 0) { 45 itt = s.begin(); 46 s.erase(itt); 47 } 48 if(s.size() == 0) break; 49 } 50 else if(b >= x){ 51 itt = s.upper_bound(a); 52 if(itt == s.begin()){ 53 if(a >= *itt) 54 s.erase(itt); 55 } 56 else{ 57 itt--; 58 s.erase(itt); 59 } 60 if(s.size() == 0) break; 61 itt = s.upper_bound(c); 62 if(itt == s.begin()){ 63 if(c >= *itt) s.erase(itt); 64 } 65 else{ 66 itt--; 67 s.erase(itt); 68 } 69 if(s.size() == 0) break; 70 } 71 else if(c >= x){ 72 int lb = 0,ub = 0; 73 itt = s.upper_bound(a); 74 if(itt == s.begin()){ 75 if(a >= *itt){ 76 s.erase(itt); 77 lb = 1; 78 } 79 } 80 else{ 81 itt--; 82 lb = 1; 83 s.erase(itt); 84 } 85 if(s.size() == 0) break; 86 itt = s.upper_bound(b); 87 if(itt == s.begin()){ 88 if(b >= *itt){ 89 s.erase(itt); 90 ub = 1; 91 } 92 } 93 else{ 94 itt--; 95 ub = 1; 96 s.erase(itt); 97 } 98 if(s.size() == 0) break; 99 if(lb == 0 && ub == 0){ 100 itt = s.upper_bound(a+b); 101 if(itt == s.begin()){ 102 if((a+b) >= *itt){ 103 s.erase(itt); 104 } 105 } 106 else{ 107 itt--; 108 s.erase(itt); 109 } 110 if(s.size() == 0) break; 111 } 112 113 /* printf("---debug--\n"); 114 for(multiset<int>::iterator p = s.begin();p != s.end();p++){ 115 printf("%d ",*p); 116 } 117 printf("---end---\n");*/ 118 119 } 120 else if((a+b) >= x){ 121 itt = s.upper_bound(c); 122 if(itt == s.begin()){ 123 if(c >= *itt){ 124 s.erase(itt); 125 } 126 } 127 else{ 128 itt--; 129 s.erase(itt); 130 } 131 if(s.size() == 0) break; 132 133 } 134 else if((a+c) >= x){ 135 itt = s.upper_bound(b); 136 if(itt == s.begin()){ 137 if(b >= *itt){ 138 s.erase(itt); 139 } 140 } 141 else{ 142 itt--; 143 s.erase(itt); 144 } 145 if(s.size() == 0) break; 146 } 147 else if((b+c) >= x){ 148 itt = s.upper_bound(a); 149 if(itt == s.begin()){ 150 if(a >= *itt){ 151 s.erase(itt); 152 } 153 } 154 else{ 155 itt--; 156 s.erase(itt); 157 } 158 if(s.size() == 0) break; 159 } 160 else{ 161 int y = 1; 162 } 163 /* printf("s.size() = %d cc = %d\n",s.size(),cc); 164 for(multiset<int>::iterator p = s.begin();p != s.end();p++){ 165 printf("%d ",*p); 166 } 167 printf("----\n");*/ 168 } 169 printf("%d\n",cc); 170 } 171 172 int main(){ 173 // freopen("in.txt","r",stdin); 174 //freopen("out.txt","w",stdout); 175 while(scanf("%d",&n) != EOF){ 176 scanf("%d %d %d",&a,&b,&c); 177 for(int i = 1;i <= n;i++) scanf("%d",&t[i]); 178 solve(); 179 } 180 return 0; 181 }
hdu 1024
将 n 个数 分成 连续的m 段，问这m段最大的和是多少
不会做---
dp[i][j] 表示前 i 个数 分成 j 段 能够得到最大的和(选取第 j 个数)
然后 dp[i][j] = max(dp[i-1][j-1] + a[i],dp[k][j-1] + a[i]) (1 <= k <= i)
看的题解做的--
但是wa 了好久---
是 因为直接输出 的dp[n] (按照这个dp的定义，必须选取第n个数)，应该再和没有选取第n个数的取一下最大值
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 #define INF 0x7fffffff 8 typedef long long LL; 9 //const long long INF = (1<<62)-1; 10 const int maxn = 1e6+5; 11 int n,m; 12 int a[maxn],b[maxn],dp[maxn]; 13 14 void solve(){ 15 int tmp; 16 memset(dp,0,sizeof(dp)); 17 memset(b,0,sizeof(b)); 18 for(int i = 1;i <= m;i++){ 19 tmp = -INF; 20 for(int j = i;j <= n;j++){ 21 dp[j] = max(dp[j-1]+a[j],b[j-1]+a[j]); 22 b[j-1] = tmp; 23 tmp = max(tmp,dp[j]); 24 // printf("dp[%d] = %d\n",j,dp[j]); 25 } 26 } 27 dp[n] = max(dp[n],tmp); 28 printf("%d\n",dp[n]); 29 } 30 31 int main(){ 32 while(scanf("%d %d",&m,&n) != EOF){ 33 for(int i = 1;i <= n;i++) scanf("%d",&a[i]); 34 solve(); 35 } 36 return 0; 37 }
hdu 1029
用map统计个数 --（话说为什么会出现在 dp 的分类里面-----）
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<map> 6 using namespace std; 7 8 typedef long long LL; 9 const int maxn = 1e6+5; 10 int a[maxn]; 11 int n; 12 13 void solve(){ 14 map<int,int> h; 15 int c = (n+1)/2; 16 for(int i = 1;i <= n;i++){ 17 h[a[i]]++; 18 if(h[a[i]] >= c){ 19 printf("%d\n",a[i]); 20 return; 21 } 22 } 23 } 24 25 int main(){ 26 while(scanf("%d",&n) != EOF){ 27 for(int i = 1;i <= n;i++) scanf("%d",&a[i]); 28 solve(); 29 } 30 return 0; 31 }
1.14
cf 294b
http://codeforces.com/problemset/problem/294/B
暴力枚举两种厚度的书分别垂直放了多少本---
dp 的做法没有想出来---再补叭-
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 105; 8 struct node{ 9 int t,w; 10 }; 11 12 node a[maxn],b[maxn]; 13 int n; 14 15 int cmp(node n1,node n2){ 16 return n1.w > n2.w; 17 } 18 19 int c1,c2; 20 21 void solve(){ 22 if(c1 != 0) sort(a+1,a+c1+1,cmp); 23 if(c2 != 0) sort(b+1,b+c2+1,cmp); 24 int ans = (1<<30)-1; 25 for(int i = 0;i <= c1;i++){ 26 for(int j = 0;j <= c2;j++){ 27 int lb = i+2*j; 28 int ub = 0; 29 for(int p = i+1;p <= c1;p++){ 30 ub += a[p].w; 31 } 32 for(int q = j+1;q <= c2;q++){ 33 ub += b[q].w; 34 } 35 if(lb >= ub){ 36 ans = min(ans,lb); 37 } 38 } 39 } 40 printf("%d\n",ans); 41 } 42 43 int main(){ 44 while(scanf("%d",&n) != EOF){ 45 c1 = 0; 46 c2 = 0; 47 int x,y; 48 for(int i = 1;i <= n;i++){ 49 scanf("%d %d",&x,&y); 50 if(x == 1) { 51 a[++c1].t = x; 52 a[c1].w = y; 53 } 54 else{ 55 b[++c2].t = x; 56 b[c2].w = y; 57 } 58 } 59 solve(); 60 } 61 return 0; 62 }
cf 294c
http://codeforces.com/contest/294/problem/C
没想出来，看的题解
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 8 typedef long long LL; 9 const int maxn = 1000+5; 10 const int mod = 1e9+7; 11 12 int n,m; 13 LL fac[maxn+5],afac[maxn+5]; 14 int a[maxn],cnt[maxn]; 15 int vis[maxn]; 16 17 LL Q_pow(LL x,LL y){ 18 LL res = 1; 19 x %= mod; 20 while(y){ 21 if(y&1) res = res*x%mod; 22 x = x*x%mod; 23 y >>= 1; 24 } 25 return res; 26 } 27 28 void Pre(){ 29 fac[0] = 1; 30 for(int i = 1;i <= maxn;i++){ 31 fac[i] = fac[i-1]*1LL*i%mod; 32 } 33 afac[maxn] = Q_pow(fac[maxn],mod-2); 34 for(int i = maxn;i >= 1;i--) afac[i-1] = afac[i]*i%mod; 35 } 36 37 void solve(){ 38 int pos = 0; 39 LL ans = 1; 40 for(int i = 1;i <= n;){ 41 int j = i; 42 if(vis[i]){ 43 pos = i; 44 i++; 45 continue; 46 } 47 while(!vis[j] && j <= n) j++; 48 int lb = j-i; 49 ans = ans*afac[lb]%mod; 50 //printf("i = %d j = %d\n",i,j); 51 if(vis[pos] && vis[j]){ 52 ans = ans*Q_pow(2,lb-1)%mod; 53 } 54 i = j; 55 } 56 ans = ans*fac[n-m]%mod; 57 printf("%I64d\n",ans); 58 } 59 60 int main(){ 61 Pre(); 62 63 /* for(int i = 1;i <= 10;i++){ 64 printf("fac[%d] = %I64d afac[%d] = %I64d\n",i,fac[i],i,afac[i]); 65 }*/ 66 67 while(scanf("%d %d",&n,&m) != EOF){ 68 memset(vis,0,sizeof(vis)); 69 for(int i = 1;i <= m;i++) { 70 scanf("%d",&a[i]); 71 vis[a[i]] = 1; 72 } 73 solve(); 74 } 75 return 0; 76 }
1.15
今天好颓- -
被hack得好惨阿---
cf 614a
http://codeforces.com/problemset/problem/614/A
可以
p[i-1]*k > r
的时候，移项来判断是不是大于 p[i-1] > r/k
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 LL l,r,k; 10 LL p[10005]; 11 12 void solve(){ 13 p[0] = 1; 14 int c = 0; 15 if(l == 1){ 16 printf("1 "); 17 c++; 18 } 19 for(int i = 1;i < 100;i++){ 20 LL tmp = 1LL*r/k; 21 if(p[i-1] > tmp) break; 22 p[i] = p[i-1]*k; 23 if(p[i] >= l && p[i] <= r) { 24 printf("%I64d ",p[i]); 25 c++; 26 } 27 } 28 if(c == 0) puts("-1"); 29 } 30 31 int main(){ 32 while(scanf("%I64d %I64d %I64d",&l,&r,&k) != EOF){ 33 solve(); 34 } 35 return 0; 36 }
另外是一神教的办法，两边取一下对数
log(p[i-1]) + log(k) > log(r)
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 #include<vector> 7 using namespace std; 8 9 typedef long long LL; 10 LL l,r,k; 11 LL p[10005]; 12 13 void solve(){ 14 p[0] = 1; 15 int c = 0; 16 if(l == 1){ 17 printf("1 "); 18 c++; 19 } 20 for(int i = 1;i < 100;i++){ 21 double lb = log(p[i-1]) + log(k); 22 double ub = log(r); 23 // printf("i = %d lb = %lf ub = %lf\n",i,lb,ub); 24 double cha = lb - ub; 25 if(cha > 1e-7 ) { 26 // printf("lb = %lf ub = %lf cha = %lf\n",lb,ub,cha); 27 break; 28 } 29 p[i] = p[i-1]*k; 30 //printf("p[%d] = %I64d ---",i,p[i]); 31 if(p[i] >= l && p[i] <= r) { 32 printf("%I64d ",p[i]); 33 c++; 34 } 35 } 36 if(c == 0) puts("-1"); 37 } 38 39 int main(){ 40 while(scanf("%I64d %I64d %I64d",&l,&r,&k) != EOF){ 41 solve(); 42 } 43 return 0; 44 }
cf 614c
http://codeforces.com/contest/614/problem/C
貌似求个圆环
然后 r = min(点到线段的距离)
R = max(P 到 点的距离)
不知道了.....
还是去抄了一个板-
1 #include <cstdio> 2 #include <algorithm> 3 #include <cmath> 4 #include <vector> 5 using namespace std; 6 //lrj计算几何模板 7 struct Point 8 { 9 double x, y; 10 Point(double x=0, double y=0) :x(x),y(y) {} 11 }; 12 typedef Point Vector; 13 14 Point read_point(void) 15 { 16 double x, y; 17 scanf("%lf%lf", &x, &y); 18 return Point(x, y); 19 } 20 21 const double EPS = 1e-10; 22 23 //向量+向量=向量 点+向量=点 24 Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } 25 26 //向量-向量=向量 点-点=向量 27 Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } 28 29 //向量*数=向量 30 Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } 31 32 //向量/数=向量 33 Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } 34 35 bool operator < (const Point& a, const Point& b) 36 { return a.x < b.x || (a.x == b.x && a.y < b.y); } 37 38 int dcmp(double x) 39 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; } 40 41 bool operator == (const Point& a, const Point& b) 42 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } 43 44 /**********************基本运算**********************/ 45 46 //点积 47 double Dot(Vector A, Vector B) 48 { return A.x*B.x + A.y*B.y; } 49 //向量的模 50 double Length(Vector A) { return sqrt(Dot(A, A)); } 51 52 //向量的夹角，返回值为弧度 53 double Angle(Vector A, Vector B) 54 { return acos(Dot(A, B) / Length(A) / Length(B)); } 55 56 //叉积 57 double Cross(Vector A, Vector B) 58 { return A.x*B.y - A.y*B.x; } 59 60 //向量AB叉乘AC的有向面积 61 double Area2(Point A, Point B, Point C) 62 { return Cross(B-A, C-A); } 63 64 //向量A旋转rad弧度 65 Vector VRotate(Vector A, double rad) 66 { 67 return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); 68 } 69 70 //将B点绕A点旋转rad弧度 71 Point PRotate(Point A, Point B, double rad) 72 { 73 return A + VRotate(B-A, rad); 74 } 75 76 //求向量A向左旋转90°的单位法向量,调用前确保A不是零向量 77 Vector Normal(Vector A) 78 { 79 double l = Length(A); 80 return Vector(-A.y/l, A.x/l); 81 } 82 83 /**********************点和直线**********************/ 84 85 //求直线P + tv 和 Q + tw的交点，调用前要确保两条直线有唯一交点 86 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) 87 { 88 Vector u = P - Q; 89 double t = Cross(w, u) / Cross(v, w); 90 return P + v*t; 91 }//在精度要求极高的情况下，可以自定义分数类 92 93 //P点到直线AB的距离 94 double DistanceToLine(Point P, Point A, Point B) 95 { 96 Vector v1 = B - A, v2 = P - A; 97 return fabs(Cross(v1, v2)) / Length(v1); //不加绝对值是有向距离 98 } 99 100 //点到线段的距离 101 double DistanceToSegment(Point P, Point A, Point B) 102 { 103 if(A == B) return Length(P - A); 104 Vector v1 = B - A, v2 = P - A, v3 = P - B; 105 if(dcmp(Dot(v1, v2)) < 0) return Length(v2); 106 else if(dcmp(Dot(v1, v3)) > 0) return Length(v3); 107 else return fabs(Cross(v1, v2)) / Length(v1); 108 } 109 110 //点在直线上的射影 111 Point GetLineProjection(Point P, Point A, Point B) 112 { 113 Vector v = B - A; 114 return A + v * (Dot(v, P - A) / Dot(v, v)); 115 } 116 117 //线段“规范”相交判定 118 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) 119 { 120 double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1); 121 double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); 122 return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0; 123 } 124 125 //判断点是否在线段上 126 bool OnSegment(Point P, Point a1, Point a2) 127 { 128 Vector v1 = a1 - P, v2 = a2 - P; 129 return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0; 130 } 131 132 //求多边形面积 133 double PolygonArea(Point* P, int n) 134 { 135 double ans = 0.0; 136 for(int i = 1; i < n - 1; ++i) 137 ans += Cross(P[i]-P[0], P[i+1]-P[0]); 138 return ans/2; 139 } 140 141 int n; 142 const int maxn = 1e6+5; 143 144 Point P; 145 Point t[maxn]; 146 147 void solve(){ 148 double r = 1.0*1e9; 149 double R = -1.0*1e9; 150 for(int i = 0;i < n;i++){ 151 double dis = sqrt((t[i].x-P.x)*(t[i].x - P.x) + (t[i].y - P.y)*(t[i].y - P.y)); 152 R = max(R,dis); 153 double dist = DistanceToSegment(P,t[i],t[(i+1)%n]); 154 r = min(r,dist); 155 //printf("R = %lf r = %lf\n",R,r); 156 } 157 double ans = 3.14159265357*(R*R-r*r); 158 printf("%.9lf\n",ans); 159 } 160 161 int main(){ 162 while(scanf("%d",&n) != EOF){ 163 scanf("%lf %lf",&P.x,&P.y); 164 for(int i = 0;i < n;i++){ 165 scanf("%lf %lf",&t[i].x,&t[i].y); 166 } 167 solve(); 168 } 169 return 0; 170 }
1.16
神经兮兮地在计蒜客看了一下午python
有一道题目是这样的--
输入 A ,B,C
再输出 A+B+C
为什么用它说的 raw_input()不行阿---
后来从 cf 扒了一个输入才通过-----诶--------------
---------------昏割线-----------------
cf 313 c
http://codeforces.com/contest/313/problem/C
先排序，再统计能够整除4的幂的数 的次数，把次数排序后，和原数组相乘
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 = 5e6+5; 9 int p[maxn],a[maxn],b[maxn]; 10 int n; 11 12 void solve(){ 13 int k; 14 for(int i = 0;i <= 14;i++){ 15 if(n == p[i]){ 16 k = i; 17 break; 18 } 19 } 20 // printf("k = %d\n",k); 21 LL ans = 0; 22 sort(a+1,a+n+1); 23 memset(b,0,sizeof(b)); 24 for(int i = 1;i <= n;i++){ 25 for(int j = 0;j <= k;j++){ 26 if(i%p[j] == 0) b[i]++; 27 } 28 } 29 sort(b+1,b+n+1); 30 for(int i = 1;i <= n;i++){ 31 ans += 1LL*a[i]*b[i]; 32 } 33 printf("%I64d\n",ans); 34 } 35 36 int main(){ 37 p[0] = 1; 38 for(int i = 1;i <= 14;i++) { 39 p[i] = p[i-1]*4; 40 // printf("p[%d] = %d\n",i,p[i]); 41 } 42 while(scanf("%d",&n) != EOF){ 43 // printf("n = %d\n",n); 44 for(int i = 1;i <= n;i++){ 45 scanf("%d",&a[i]); 46 // printf("a[%d] = %d\n",i,a[i]); 47 } 48 solve(); 49 } 50 return 0; 51 }
1.17
cf 289c
http://codeforces.com/problemset/problem/289/C
分 n-k 的奇偶构造一下
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 1e6+5; 8 char s[maxn]; 9 int n,k; 10 11 12 void solve(){ 13 char t[maxn]; 14 if(n < k){ 15 puts("-1"); 16 return; 17 } 18 if(n != 1 && k == 1){ 19 puts("-1"); 20 return; 21 } 22 23 if(n == k){ 24 for(int i = 0;i < n;i++){ 25 char c = 'a'+i; 26 printf("%c",c); 27 } 28 printf("\n"); 29 return; 30 } 31 if(k == 2){ 32 for(int i = 1;i <= n;i++){ 33 if(i%2) printf("a"); 34 else printf("b"); 35 } 36 printf("\n"); 37 return; 38 } 39 40 if((n-k)%2 == 0){ 41 for(int i = 1;i <= n-k;i++){ 42 if(i%2 == 1) t[i] = 'a'; 43 else t[i] = 'b'; 44 } 45 for(int i = n-k+1;i <= n;i++){ 46 char c = 'a' + i-(n-k+1); 47 t[i] = c; 48 } 49 } 50 else{ 51 k--; 52 for(int i = 1;i <= n-k;i++){ 53 if(i%2 == 1) t[i] = 'a'; 54 else t[i] = 'b'; 55 } 56 for(int i = n-k+1;i <= n;i++){ 57 int z = i-(n-k+1); 58 if(z == 0) t[i] = 'a'; 59 else{ 60 z++; 61 t[i] = 'a'+z; 62 } 63 } 64 } 65 for(int i = 1;i <= n;i++) printf("%c",t[i]); 66 // printf("%s",t+1); 67 printf("\n"); 68 } 69 70 int main(){ 71 while(scanf("%d %d",&n,&k) != EOF){ 72 solve(); 73 } 74 return 0; 75 }
cf 614d
http://codeforces.com/problemset/problem/614/D
先对a数组从小到大排序后，枚举达到 A 的个数，（a值大的优先），
再二分剩下的，使得前 x 个的值一样（前x个这时候是作为最小值）
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 = 1e6+5; 10 int a[maxn]; 11 LL p[maxn]; 12 int n,A,cf,cm; 13 LL m,y; 14 LL sum[maxn]; 15 LL aa[maxn]; 16 17 int ok(int x){ 18 return y >= p[x]; 19 } 20 21 struct node{ 22 int u; 23 int id; 24 }q[maxn]; 25 26 int cmp(node n1,node n2){ 27 return n1.u < n2.u; 28 } 29 30 int cmp0(node n1,node n2){ 31 return n1.id < n2.id; 32 } 33 34 void solve(){ 35 memset(p,0,sizeof(p)); 36 sort(a+1,a+n+1); 37 p[1] = 0; 38 for(int i = 2;i <= n;i++){ 39 p[i] = p[i-1] + 1LL*(i-1)*(a[i]-a[i-1]); 40 // printf("p[%d] = %d\n",i,p[i]); 41 } 42 for(int i = 1;i <= n;i++) aa[i] = a[n-i+1]; 43 sum[0] = 0; 44 for(int i = 1;i <= n;i++) { 45 sum[i] = sum[i-1] + 1LL*aa[i]; 46 // printf("sum[%d] = %I64d\n",i,sum[i]); 47 } 48 49 int lb,ub,mid; 50 LL ans = 0; 51 LL tmp; 52 int flag = 0; 53 int ed = 0,edd = 0,cnt = 0; 54 for(int i = 0;i <= n;i++){ 55 lb = 0; 56 ub = n-i; 57 y = m-1LL*i*A + sum[i]; 58 if(y < 0) continue; 59 tmp = 1LL*i*cf; 60 if(i == n){ 61 tmp += 1LL*cm*A; 62 flag = 1; 63 ans = max(ans,tmp); 64 // printf("----tmp = %d\n",tmp); 65 continue; 66 } 67 int pos = -1; 68 for(int k = 0;k <= 50;k++){ 69 mid = (lb+ub)/2; 70 if(ok(mid)){ 71 lb = mid+1; 72 pos = max(pos,mid); 73 } 74 else ub = mid; 75 // printf("i = %d lb = %d ub = %d mid = %d\n",i,lb,ub,mid); 76 } 77 int minn = min((y-p[pos])/pos + a[pos],1LL*A); 78 tmp += 1LL*minn*cm; 79 if(tmp > ans) ed = minn,edd = pos,cnt = i; 80 ans = max(ans,tmp); 81 // printf("i = %d pos = %d y = %I64d tmp = %I64d\n",i,pos,y,tmp); 82 } 83 printf("%I64d\n",ans); 84 if(flag){ 85 int w = A; 86 for(int i = 1;i <= n;i++) printf("%d ",w); 87 printf("\n"); 88 return; 89 } 90 sort(q+1,q+n+1,cmp); 91 for(int i = 1;i <= edd;i++) q[i].u = ed; 92 for(int i = n;i >= 1;i--){ 93 if(cnt) q[i].u = A,cnt--; 94 if(cnt == 0) break; 95 } 96 sort(q+1,q+n+1,cmp0); 97 for(int i = 1;i <= n;i++) printf("%d ",q[i].u); 98 printf("\n"); 99 } 100 101 int main(){ 102 while(scanf("%d %d %d %d %I64d",&n,&A,&cf,&cm,&m) != EOF){ 103 for(int i = 1;i <= n;i++){ 104 scanf("%d",&a[i]); 105 q[i].u = a[i]; 106 q[i].id = i; 107 } 108 solve(); 109 } 110 return 0; 111 }