标签:二分 lines desc nes closed tor code form div
Description
Input
Output
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
首先开一个数组枚举cd和,然后通过枚举ab和在新的数组里二分查找解
#include <cstdio> #include <cmath> #include <cctype> #include <iostream> #include <cstring> #include <algorithm> #include <string> #include <stack> #include <vector> #include <map> #include <set> using namespace std; typedef long long LL; string s; LL a[4005],b[4005],c[4005],d[4005]; LL cd[16000001] = {0}; LL n,res = 0; int main() { // freopen("test.in","r",stdin); ios::sync_with_stdio(false); cin >> n; for (int i=0;i<n;i++){ cin >> a[i] >> b[i] >> c[i] >> d[i]; } for (int i=0;i<n;i++){ for (int j=0;j<n;j++){ cd[i*n+j] = c[i] + d[j]; } } sort(cd,cd+n*n); for (int i=0;i<n;i++){ for (int j=0;j<n;j++){ LL now = a[i] + b[j]; LL need = 0 - now; res += upper_bound(cd,cd+n*n,need) - lower_bound(cd,cd+n*n,need); } } cout << res; return 0; }
poj 2785 4 Values whose Sum is 0
标签:二分 lines desc nes closed tor code form div
原文地址:http://www.cnblogs.com/ToTOrz/p/7270291.html