4 Values whose Sum is 0
Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions: 27860 | Accepted: 8383 | |
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 2
28 ) 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
#include <iostream>
#include <algorithm>
using namespace std;
const int mn = 4010;
int a[4][mn];
int b[mn * mn];
int c[mn * mn];
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < 4; j++)
cin >> a[j][i];
// 枚举上两行和下两行各自的可能性 再二分查找
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
b[i * n + j] = a[0][i] + a[1][j];
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
c[i * n + j] = a[2][i] + a[3][j];
}
long long res = 0;
sort(c, c + n * n);
for (int i = 0; i < n * n; i++)
{
int u = -b[i];
res += upper_bound(c, c + n * n, u) - lower_bound(c, c + n * n, u);
// (>的标号) - (≥ 的标号) = =的个数
}
cout << res << endl;
return 0;
}