HDU 5213 Lucky 莫队+容斥

Lucky

Problem Description

 

WLD is always very lucky.His secret is a lucky number K.k is a fixed odd number. Now he meets a stranger with N numbers:a1,a2,...,aN.The stranger asks him M questions.Each question is like this:Given two ranges [Li,Ri] and [Ui,Vi],you can choose two numbers X and Y to make aX+aY=K.The X you can choose is between Li and Ri and the Y you can choose is between Ui and Vi.How many pairs of numbers(X,Y) you can choose?
If WLD can answer all the questions correctly,he'll be the luckiest man in the world.Can you help him?

 

 

Input

 

There are multiple cases.(At MOST 5)

For each case:

The first line contains an integer N(1≤N≤30000).

The following line contains an integer K(2≤K≤2∗N),WLD's lucky number.K is odd.

The following line contains N integers a1,a2,...,aN(1≤ai≤N).

The following line contains an integer M(1≤M≤30000),the sum of the questions WLD has to answer.

The following M lines,the i-th line contains 4 numbers Li,Ri,Ui,Vi(1≤Li≤Ri<Ui≤Vi≤N),describing the i-th question the stranger asks.

 

 

Output

 

For each case:

Print the total of pairs WLD can choose for each question.

 

 

Sample Input

 

5 3 1 2 1 2 3 1 1 2 3 5

 

 

Sample Output

 

2

Hint

a1+a4=a2+a3=3=K. So we have two pairs of numbers (1,4) and (2,3). Good luck!

 

 

题意 ：

　　给你你n个数一个k

　　m次询问，每次给你两区间

　　问你这两个区间 任选两个数a[i] + a[j] = k 的对数

题解：

　　这道题需要一些莫队算法的知识 定义记号f(A,B)f(A,B)表示询问区间A，B时的答案 用记号＋表示集合的并 利用莫队算法我们可以计算出任意f(A,A)f(A,A)的值

　　不妨假设A=[l1,r1],B=[l2,r2],C=[r1+1,l2-1]A=[l1,r1],B=[l2,r2],C=[r1+1,l2−1] 

　　容易知道f(A,B)=f(A+B+C,A+B+C)+f(C,C)-f(A+C,A+C)-f(C+B,C+B)f(A,B)=f(A+B+C,A+B+C)+f(C,C)−f(A+C,A+C)−f(C+B,C+B) 

　　因此一个询问被拆成四个可以用莫队算法做的询问 总的时间复杂度为O(msqrt(n))O(msqrt(n))

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 3e4+5, M = 6e4+5, mod = 1e9+7, inf = 1e9+1000;
typedef long long ll;

int T,n,a[N],ans[N],belong[N],mp[M * 10],k;
struct ss{
    int l,r,id,res;
    ss () {}
    ss (int l,int r,int id,int res) : l(l), r(r), id(id), res(res) {}
}Q[N * 4];
bool operator < (ss s1 , ss s2) {
    if(belong[s1.l] == belong[s2.l]) return s1.r<s2.r;
    else return belong[s1.l] < belong[s2.l];
}

int main()
{
    while(~scanf("%d",&n)) {
        memset(mp,0,sizeof(mp));
        memset(ans,0,sizeof(ans));
        scanf("%d",&k);
        for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
        int q,cnt=0;cin>>q;
        for(int i = 1; i <= q; ++i) {
            int l1,r1,l2,r2;
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            Q[++cnt] = ss (l1,r2,i,1);
            if(l2-1>=r1+1)Q[++cnt] = ss (r1+1,l2-1,i,1);
            Q[++cnt] = ss(r1+1,r2,i,-1);
            Q[++cnt] = ss(l1,l2-1,i,-1);
        }
        int t = sqrt(n);
        for(int i = 1; i <= n; ++i) belong[i] = (i-1) / t + 1;
        sort(Q+1,Q+cnt+1);
        int l = 1, r = 0, ret = 0;
        for(int i = 1; i <= cnt; ++i) {
            for(;r<Q[i].r;r++) {
                ret += mp[k-a[r+1]+M];
                mp[a[r+1]+M]++;
            }
            for(;l>Q[i].l;l--) {
                ret += mp[k-a[l-1]+M];
                mp[a[l-1]+M]++;
            }
            for(;r>Q[i].r;r--) {
                mp[a[r]+M]--;
                ret -= mp[k-a[r]+M];

            }
            for(;l<Q[i].l;l++) {
                mp[a[l]+M]--;
                ret -= mp[k-a[l]+M];

            }
           // cout<<Q[i].l<<" "<<Q[i].r<<" ";
            //cout<<Q[i].res*ret<<endl;
            ans[Q[i].id] += Q[i].res*ret;
        }
        for(int i = 1; i <= q; ++i) {
            printf("%d\n",ans[i]);
        }
    }
}