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
首先把a+b+c+d = 0化为a+b = -c-d
然后利用二分进行优化，再利用Lower bound和upper bound的差来统计符合的答案个数，刚开始没有想到，只是求解是否存在，没考虑有相同数的情况
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 typedef long long ll;
7 int n, a[4005][5], ab[4005*4005];
8 int main(void)
9 {
10 while(cin >> n)
11 {
12 for(int i = 0; i < n; i++)
13 for(int j = 0; j < 4; j++)
14 scanf("%d", &a[i][j]);
15 for(int i = 0; i < n; i++)
16 for(int j = 0; j < n; j++)
17 ab[i*n+j] = a[i][0]+a[j][1];
18 ll ans = 0;
19 sort(ab, ab+n*n);
20 for(int i = 0; i < n; i++)
21 for(int j = 0; j < n; j++)
22 {
23 int del = -(a[i][2]+a[j][3]);
24 ans += upper_bound(ab, ab+n*n, del)-lower_bound(ab, ab+n*n, del);
25 }
26 printf("%lld\n", ans);
27 }
28 return 0;
29 }
4 Values whose Sum is 0
原文：https://www.cnblogs.com/fhzy291146030/p/9371127.html
