2019 杭电多校（第五场）
1002 three arrays
http://acm.hdu.edu.cn/showproblem.php?pid=6625
题意
给你两个数组 让你排序 使他们的对应异或值字典序最小
思路
https://www.bilibili.com/video/av62396512?from=search&seid=2029202226881211707
代码
（调自闭了 请队友帮的忙（01字典树多跑了一位 初始化函数没有调用))
#include
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
int a[maxn],b[maxn];
int tree[50*maxn][2],tree2[50*maxn][2];
int num[50*maxn],num2[50*maxn];
int tot,tot2,top;
int n;
int ans[maxn];
int qw[40];
void init()
{
tot = tot2 = 1;
top = 0;
tree[0][0] = tree[0][1] = tree2[0][1] = tree2[0][0] = 0;
}
void add(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(tree[u][v] == 0)
{
tree[tot][0] = tree[tot][1] = 0;
num[tot] = 0;
tree[u][v] = tot++;
}
u = tree[u][v];
num[u]++;
}
}
void add2(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(tree2[u][v] == 0)
{
tree2[tot2][0] = tree2[tot2][1] = 0;
num2[tot2] = 0;
tree2[u][v] = tot2++;
}
u = tree2[u][v];
num2[u]++;
}
}
void dfs(int x1,int x2,int len,int res,int sum)
{
if(len == 31)
{
for(int i = 1;i <= sum;i++)
{
ans[++top] = res;
}
return ;
}
int l = num[tree[x1][0]],r = num[tree[x1][1]];
int ll = num2[tree2[x2][0]],rr = num2[tree2[x2][1]];
if(l>0&&ll>0)
{
int mi = min(l,ll);
mi = min(mi,sum);
l -= mi,ll -= mi;
dfs(tree[x1][0],tree2[x2][0],len+1,res,mi);
num[tree[x1][0]] -= mi,num2[tree2[x2][0]] -= mi;
}
if(r>0&&rr>0)
{
int mi = min(r,rr);
mi = min(mi,sum);
r -= mi,rr -= mi;
dfs(tree[x1][1],tree2[x2][1],len+1,res,mi);
num[tree[x1][1]] -= mi,num2[tree2[x2][1]] -= mi;
}
if(l>0&&rr>0)
{
int mi = min(l,rr);
mi = min(mi,sum);
l -= mi,rr -= mi;
dfs(tree[x1][0],tree2[x2][1],len+1,res^qw[29-len+1],mi);
num[tree[x1][0]] -= mi,num2[tree2[x2][1]] -= mi;
}
if(r>0&&ll>0)
{
int mi = min(r,ll);
mi = min(mi,sum);
r -= mi,ll -= mi;
dfs(tree[x1][1],tree2[x2][0],len+1,res^qw[29-len+1],mi);
num[tree[x1][1]] -= mi,num2[tree2[x2][0]] -= mi;
}
}
int main()
{
int T;
scanf("%d",&T);
qw[0] = 1;
for(int i = 1;i <= 30;i++)
{
qw[i] = qw[i-1]*2;
}
while(T--)
{
init();
int n;
scanf("%d",&n);
for(int i = 1;i <= n;i++)
{
scanf("%d",&a[i]);
add(a[i]);
}
for(int i = 1;i <= n;i++)
{
scanf("%d",&b[i]);
add2(b[i]);
}
dfs(0,0,1,0,n);
sort(ans+1,ans+1+n);
for(int i = 1;i <= n;i++)
{
if(i != 1) printf(" ");
printf("%d",ans[i]);
}
printf("\n");
}
return 0;
}
做法二
#include
using namespace std;
const int maxn = 1e5+100;
int tree[maxn*40][2],tree2[maxn*40][2];
int num[maxn*40],num2[maxn*40];
int val[maxn*40],val2[maxn*40];
int tot,tot2,top;
int a[maxn],b[maxn];
map mp;
int n;
int ans[maxn];
void init()
{
tot = tot2 = 1;
top = 0;
tree[0][0] = tree[0][1] = tree2[0][0] = tree2[0][1] = 0;
mp.clear();
}
void add(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(tree[u][v] == 0)
{
tree[tot][0] = tree[tot][1] = 0;
num[tot] = 0;
tree[u][v] = tot++;
}
u = tree[u][v];
num[u]++;
}
val[u] = x;
}
void add2(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(tree2[u][v] == 0)
{
tree2[tot2][0] = tree2[tot2][1] = 0;
num2[tot2] = 0;
tree2[u][v] = tot2++;
}
u = tree2[u][v];
num2[u]++;
}
val2[u] = x;
}
int query(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(num[tree[u][v]])
{
u = tree[u][v];
}
else u = tree[u][v^1];
}
return val[u];
}
int query2(int x)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
if(num2[tree2[u][v]])
{
u = tree2[u][v];
}
else u = tree2[u][v^1];
}
return val2[u];
}
int updata(int x,int y)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
u = tree[u][v];
num[u]+=y;
}
}
int updata2(int x,int y)
{
int u = 0;
for(int i = 29;i >= 0;i--)
{
int v = (x>>i)&1;
u = tree2[u][v];
num2[u]+=y;
}
}
int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
init();
cin>>n;
for(int i = 1;i <= n;i++)
{
cin>>a[i];
add(a[i]);
mp[a[i]]++;
}
for(int i = 1;i <= n;i++)
{
cin>>b[i];
add2(b[i]);
}
int p = 1;
int aa,bb,cc;
while(p <= n)
{
if(mp[a[p]] == 0)
{
p++;
if(p > n) break;
continue;
}
aa = a[p];
bb = query2(aa);
cc = query(bb);
while(cc != aa)
{
aa = cc;
bb = query2(aa);
cc = query(bb);
}
ans[++top] = aa^bb;
updata(aa,-1);
updata2(bb,-1);
mp[aa]--;
}
sort(ans+1,ans+1+n);
for(int i = 1;i <= n;i++)
{
if(i != 1) printf(" ");
printf("%d",ans[i]);
}
printf("\n");
}
return 0;
}
1004 equation （数学）
http://acm.hdu.edu.cn/showproblem.php?pid=6627
题意
给你一个方程 求解
思路
对于绝对值 不同的区间去绝对值后式子不同 即会有（n+1）个求解区间 分别求解即可
代码
#include
using namespace std;
const int maxn = 1e5+100;
struct node
{
int a,b;
double x;
int id;
}qw[maxn],ans[maxn];
int a[maxn],b[maxn];
int n, c, A, B;
double l,r;
queue q;
int f;
bool cmp(node a,node b)
{
if(a.x != b.x) return a.x < b.x;
else return a.id < b.id;
}
void solve()
{
if(A==0&&B == c)
{
f = 1;
return ;
}
double ans;
ans = (double)(c-B)/A;
if(ans > l&&ans <= r)
{
int as = 1;
if(ans < 0) as = -1;
int x = (c - B),y = A;
x = abs(x),y = abs(y);
int z;
if(x == 0) z = y;
else z = __gcd(x,y);
node u;
u.a = x / z * as,u.b = y / z;
q.push(u);
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&c);
A = 0,B = 0;
for(int i = 1;i <= n;i++)
{
scanf("%d%d",&a[i],&b[i]);
qw[i].a = a[i],qw[i].b = b[i];
qw[i].id = i;
qw[i].x = (double)b[i]/a[i]*(-1.0);
A += -a[i],B += -b[i];
}
sort(qw+1,qw+1+n,cmp);
f = 0;
while(!q.empty()) q.pop();
l = -1e18,r = -1e18;
qw[n+1].x = 1e18;
for(int i = 1;i <= n+1;i++)
{
l = r,r = qw[i].x;
solve();
A += qw[i].a*2;
B += qw[i].b*2;
}
if(f) printf("-1\n");
else
{
printf("%d",q.size());
while(!q.empty())
{
node u = q.front();
q.pop();
printf(" %d/%d",u.a,u.b);
}
printf("\n");
}
}
return 0;
}
1005 permutation 1 （思维）
http://acm.hdu.edu.cn/showproblem.php?pid=6628
题意
给你n个数 1 到 n 问你全排列相邻差字典序第k大的是哪个
思路
k最大10000 8个数即可 n小于8位之间暴力 大于8位直接填最后8位之前 最大8位暴力
#include
using namespace std;
typedef long long ll;
struct node
{
int a[30];
} qw[1000000];
int vis[30];
int p,n,m;
int xx[30];
bool cmp(node a,node b)
{
for(int i = 2; i <= m; i++)
{
if(a.a[i]-a.a[i-1]!= b.a[i] - b.a[i-1])
{
return a.a[i]-a.a[i-1] < b.a[i] - b.a[i-1];
}
}
}
void dfs(int x)
{
if(x == m+1)
{
p++;
for(int i = 1; i <= m; i++)
{
qw[p].a[i] = xx[i];
}
return ;
}
for(int i = 1; i <= n; i++)
{
if(vis[i] == 0)
{
vis[i] = 1;
xx[x] = i;
dfs(x+1);
vis[i] = 0;
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int k;
scanf("%d%d",&n,&k);
m = n;
p = 0;
memset(vis,0,sizeof(vis));
if(n <= 8)
{
dfs(1);
sort(qw+1,qw+1+p,cmp);
for(int i = 1; i <= n; i++)
{
if(i != 1)
printf(" ");
printf("%d",qw[k].a[i]);
}
printf("\n");
}
else
{
printf("%d",n);
vis[n] =1;
for(int i = 1; i <= n - 8 - 1; i++)
{
printf(" %d",i);
vis[i] = 1;
xx[1] = i;
}
m = 9;
dfs(2);
sort(qw+1,qw+1+p,cmp);
for(int i = 2; i <= m; i++)
{
printf(" %d",qw[k].a[i]);
}
printf("\n");
}
}
return 0;
} 1006 string matching （扩展kmp）
http://acm.hdu.edu.cn/showproblem.php?pid=6629
题意
给你一个字符串 问从第2位 每一位和字符串相等前缀多长 （比较多少次 直到完全匹配 或 失败）
思路
拓展kmp 注意完全匹配时不需要加一次失败匹配
代码
#include
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
typedef long long ll;
char s[N];
int nex[N],extend[N];
ll sum;
void get_next()
{
int len = strlen(s);
nex[0] = len;
int mx = 0,id;
for(int i = 1;i < len;i++)
{
if(i < mx) nex[i] = min(mx-i,nex[i-id]);
else nex[i] = 0;
while(s[i+nex[i]] == s[nex[i]]) nex[i]++;
if(mx < i + nex[i])
{
id = i;
mx = i + nex[i];
}
sum += (ll)min(nex[i]+1,len-i);
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%s",s);
sum = 0;
get_next();
ll n = strlen(s);
printf("%lld\n",sum);
}
return 0;
} 1007 permutation 2
http://acm.hdu.edu.cn/showproblem.php?pid=6630
题意
给你n个数 1-n 给你其中两个做第一和最后 打其他数填到里面 有多少种填法
思路
很容易想到如果第一位不是1 要把1-x-1填成 1 ...... x-1
最后一位不是n n后面要填成 y-1......y
中间就是 dp[i] = dp[i-1] + dp[i-3]
代码
#include
using namespace std;
typedef long long ll;
const ll mod = 998244353;
ll dp[100005];
int main()
{
int T;
scanf("%d",&T);
dp[1] = 1;
dp[2] = 1;
dp[3] = 1;
for(int i = 4;i <= 100000;i++)
{
dp[i] = dp[i-1] + dp[i-3];
dp[i] %= mod;
}
while(T--)
{
ll n,x,y;
scanf("%lld%lld%lld",&n,&x,&y);
if(x > y) swap(x,y);
if(x != 1) x++;
if(y != n) y--;
printf("%lld\n",dp[y-x+1]);
}
return 0;
}