POJ 2785 4 Values whose Sum is 0

最新推荐文章于 2021-10-24 16:23:06 发布

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最新推荐文章于 2021-10-24 16:23:06 发布

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文章标签： POJ

本文链接：https://blog.youkuaiyun.com/qq_26658823/article/details/51979032

探讨了如何解决四数之和为零的问题，通过预处理将两组数据的组合和进行排序，并利用upper_bound与lower_bound函数快速查找匹配项。

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4 Values whose Sum is 0

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

题目大意：给出4组数据，在每一组中选一个数，使4个数的和为零，问有多少种选法。

解题思路：分别求出前两组数据与后两组数据的和，然后对它们排序，利用upper_bound与lower_bound函数可以求出和为0的组数。

lower_bound从已排好序的序列a中利用二分搜索找出指向满足ai >= k的ai的最小指针，upper_bound函数返回指向满足ai > k的ai的最小指针，所以长度为n的有序序列a中k的个数可以利用以下方法求出：

upper_bound(a,a + n,k) - lower_bound(a,a + n,k)

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>

using namespace std;
typedef long long ll;
const int maxn = 4005;

ll a[maxn],b[maxn],c[maxn],d[maxn],ab[maxn * maxn],cd[maxn * maxn];
int main()
{
	int i,j,k,l,n,ans;
	while(scanf("%d",&n) != EOF){
		for(i = 0;i < n;i++){
			scanf("%lld %lld %lld %lld",&a[i],&b[i],&c[i],&d[i]);
		}
		k = 0;
		for(i = 0;i < n;i++){
			for(j = 0;j < n;j++){
				ab[k] = a[i] + b[j];
				cd[k] = c[i] + d[j];
				k++;
			}
		}
		sort(ab,ab + k);
		sort(cd,cd + k);
		ans = 0;
		for(i = 0;i < k;i++){
			ans += upper_bound(ab,ab + k,-cd[i]) - lower_bound(ab,ab + k,-cd[i]);
		}
		printf("%d\n",ans);
	}
	return 0;
}