POJ 2785：4 Values whose Sum is 0

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).

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：

通过枚举后两列的所有可能的值，然后再用折半枚举算出所需要的ab中的和为0的值.

源代码：

#include<iostream>
#include<bits/stdc++.h>
#define N 4005using namespace std;
int a
,b
,c
,d
;
int cd[N*N],t;
long long ans;
void work()
{
ans=0;
for(int i=0;i<t;i++)
for(int j=0;j<t;j++)
cd[i*t+j]=c[i]+d[j];
sort(cd,cd+t*t);
for(int i=0;i<t;i++)
for(int j=0;j<t;j++)
{
int ab=-(a[i]+b[j]);
//upper_bound返回的是值为ab的元素可以插入的最后一个位置（上界）   （可以认为大于ab的位置）
//lower_bound返回的是键值为ab的元素可以插入的位置的第一个位置（下界）。（可以认为大于或等于ab的位置）
ans+=upper_bound(cd,cd+t*t,ab)-lower_bound(cd,cd+t*t,ab);
}
printf("%lld\n",ans);

}
int main()
{
scanf("%d",&t);
for(int i=0;i<t;i++)
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
work();
return 0;
}