4 Values whose Sum is 0 二分
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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).
前后两个数组个元素的和球出来,然后二分搜索求解。
#include<iostream> #include<algorithm> #include<cstdio> using namespace std; #define MAXN 4004 int a[MAXN], b[MAXN], c[MAXN], d[MAXN], sum1[MAXN*MAXN], sum2[MAXN*MAXN]; int main() { int n; while (scanf("%d", &n) != EOF) { for (int i = 0; i < n; i++) { scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]); } int l1, l2; l1 = l2 = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { sum1[l1++] = a[i] + b[j]; sum2[l2++] = c[i] + d[j]; } } sort(sum1, sum1 + l1); int ans = 0; for (int i = 0; i < l2; i++) { int beg = 0, end = l1 - 1; while (beg <= end) { int mid = (beg + end) / 2; if (sum1[mid]+ sum2[i]==0) { ans++; for (int p = mid + 1; p < l1; p++) if (sum1[p] + sum2[i] != 0) break; else ans++; for (int p = mid - 1; p >= 0; p--) if (sum1[p] + sum2[i] != 0) break; else ans++; break; } if (sum1[mid] + sum2[i]<0) beg = mid + 1; else end = mid - 1; } } printf("%d\n", ans); } }
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