POJ - 2785 whose Sum is 0

最新推荐文章于 2021-07-28 08:44:58 发布

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最新推荐文章于 2021-07-28 08:44:58 发布

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分类专栏：二分

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本文介绍了一个经典的编程问题——SUM问题的高效求解方法。通过预处理A和B两个列表的所有可能和，并使用二分查找来匹配C和D列表中元素组合的相反数，实现了快速寻找所有元素组合使得四数之和为零。

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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

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).

先AB的和做表，再计算CD的和，在表中二分查找对应的数，节约空间CD和没必要做表，由于这题求的是组合方式，不能去重，用的是upper_bound和lower_bound求元素的上下标，可得元素个数

#include <iostream>
#include <stdio.h>
#include<algorithm>
using namespace std;
int a[4001],b[4001],c[4001],d[4001],s1[16000005];

int main()
{
    int n,i,j,x,t;
    unsigned long long y;
    scanf("%d",&n);
    x=1,y=0;
    for(i=0;i<n;i++){
        scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            s1[x]=a[i]+b[j];
            x++;
        }
    }
    sort(s1+1,s1+x);
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            t=c[i]+d[j];
            t=0-t;
            y+=upper_bound(s1+1,s1+x,t)-lower_bound(s1+1,s1+x,t);

        }
    }

    cout<<y<<endl;
    return 0;
}