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 228 ) 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*/