4 Values whose Sum is 0
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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
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 ", ans); 27 } 28 return 0; 29 }
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