hdu5213 Lucky 莫队算法+容斥

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题目：给你N个数字，一个K，M个询问L,R,U,V，问你在区间[L,R]内找一个数x，在区间[U,V]内找一个数y，使得x+y=K，一共有多少种方法。

思路：莫队算法+容斥。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst int maxn=6e4+50;struct Ask{    int L,R,id;}ask[maxn];int ans[maxn];int arr[maxn],num[maxn];//从1开始计数int n,m,siz,K;//n个数，m个询问，块的大小LL temp;bool cmp(Ask a,Ask b){    if(a.L/siz!=b.L/siz) return a.L/siz<b.L/siz;    else return a.R<b.R;}void add(int pos){    temp+=num[K-arr[pos]];    num[arr[pos]]++;}void sub(int pos){    temp-=num[K-arr[pos]];    num[arr[pos]]--;}void solve(){    temp=0;    mm(num,0);    int L=1,R=0;    for(int i=1;i<=m;i++){        if(ask[i].L>=ask[i].R){            ans[ask[i].id]=0;            continue;        }        while(R<ask[i].R){            R++;            add(R);        }        while(R>ask[i].R){            sub(R);            R--;        }        while(L<ask[i].L){            sub(L);            L++;        }        while(L>ask[i].L){            L--;            add(L);        }        ans[ask[i].id]=temp;    }}int L[maxn],R[maxn],U[maxn],V[maxn];int ret[maxn];int main(){    while(~scanf("%d",&n)){        scanf("%d",&K);        for(int i=1;i<=n;i++)            scanf("%d",&arr[i]);        scanf("%d",&m);        for(int i=1;i<=m;i++)            scanf("%d%d%d%d",&L[i],&R[i],&U[i],&V[i]);        siz=(int)sqrt(n*1.0);        for(int i=1;i<=m;i++){            ask[i].id=i;            ask[i].L=L[i];            ask[i].R=V[i];        }        sort(ask+1,ask+m+1,cmp);        solve();        for(int i=1;i<=m;i++) ret[i]=ans[i];        for(int i=1;i<=m;i++){            ask[i].id=i;            ask[i].L=L[i];            ask[i].R=U[i]-1;        }        sort(ask+1,ask+m+1,cmp);        solve();        for(int i=1;i<=m;i++) ret[i]-=ans[i];        for(int i=1;i<=m;i++){            ask[i].id=i;            ask[i].L=R[i]+1;            ask[i].R=V[i];        }        sort(ask+1,ask+m+1,cmp);        solve();        for(int i=1;i<=m;i++) ret[i]-=ans[i];        for(int i=1;i<=m;i++){            ask[i].id=i;            ask[i].L=R[i]+1;            ask[i].R=U[i]-1;        }        sort(ask+1,ask+m+1,cmp);        solve();        for(int i=1;i<=m;i++) ret[i]+=ans[i];        for(int i=1;i<=m;i++)            printf("%d\n",ret[i]);    }    return 0;}

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