标签：2785 int Sum 30 sum2 sum1 pos ++ Values

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 .

输入描述:

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 .

输出描述:

For each input file, your program has to write the number quadruplets whose sum is zero.
示例1
输入
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
输出
5

备注:

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).

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 4010;
int a[N], b[N], c[N], d[N], sum1[N * N + 10], sum2[N * N + 10], n;
int main()
{
	cin >> n;
	for(int i = 0; i < n; i++)
		cin >> a[i] >> b[i] >> c[i] >> d[i];
	int k = 0;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++)
		{
			sum1[k] = a[i] + b[j];
			sum2[k] = c[i] + d[j];
			k++;
		}
	int ans = 0;
	sort(sum1, sum1 + k), sort(sum2, sum2 + k);
	for(int i = 0; i < k; i++)
	{
		int pos = lower_bound(sum2, sum2 + k, -sum1[i]) - sum2;
		while(sum2[pos] == -sum1[i] && pos < k)
			ans++, pos++;
	}
	cout << ans << endl;
	return 0;
}

标签：2785,int,Sum,30,sum2,sum1,pos,++,Values
来源： https://blog.csdn.net/qq_44826711/article/details/114017252

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