POJ 2785-4 Values whose Sum is 0（a+b+c+d=0-折半枚举）

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 19331		Accepted: 5783
Case Time Limit: 5000MS

Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2 ²⁸ ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

Source

Southwestern Europe 2005

题目意思：

有abcd四个数组，各抽取一个数使得四个数之和为0；问共有多少种取法。

解题思路：

先取cd中的。再用折半枚举算出所需要的ab中的和的值。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<malloc.h>
#include<algorithm>
#define MAXN 4005
#define inf 0x3f3f3f3f
using namespace std;
int n;
int a[MAXN],b[MAXN],c[MAXN],d[MAXN];
int cd[MAXN*MAXN];//c,d中数字的组合方法

void solve()
{
    for(int i=0; i<n; ++i)
        for(int j=0; j<n; ++j)
            cd[i*n+j]=c[i]+d[j];//枚举c,d数字之和的所有情况
    sort(cd,cd+n*n);
    long long res=0;
    for(int i=0; i<n; ++i)
        for(int j=0; j<n; ++j)
        {
            int addcd=-(a[i]+b[j]);//需要加上的数
            //参数：数组，数组长度，目标值
            //upper_bound返回第一个出现的位置
            //lower_bound返回最后一个的后面一个出现的位置
            //都没有就返回应该出现的位置使得sort依然有序
            res+=upper_bound(cd,cd+n*n,addcd)-lower_bound(cd,cd+n*n,addcd);
        }
    cout<<res<<endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=0; i<n; ++i)
        cin>>a[i]>>b[i]>>c[i]>>d[i];
    solve();
    return 0;
}
/*
6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45
*/