4 Values whose Sum is 0 POJ - 2785
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4 Values whose Sum is 0
Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions: 29243 | Accepted: 8887 | |
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 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).
题意:ABCD四个数列,每个数列抽一个数字,使得和为0,问一共有几组。当一个数列中有多个相同的数字时,把他们作为不同的数字看待。
思路:直接爆搜肯定会超时,将它们对半分成AB和CD再考虑就可以解决。比如在AB中取出a,b后,CD中要取出-a-b,因此先把CD中所有取数字的情况n*n种给纪录下来,然后排序,用二分查找次数。
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cstdlib> #include<queue> #include<set> #include<vector> using namespace std; #define INF 0x3f3f3f3f #define eps 1e-10 typedef long long ll; const int maxn = 4002; const int mod = 1e9 + 7; int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } int n; int a[maxn],b[maxn],c[maxn],d[maxn]; int cd[maxn*maxn]; void solve() { 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); ll res=0; for(int i=0;i<n;i++) for(int j=0;j<n;j++) { int ab=-(a[i]+b[j]); res+=upper_bound(cd,cd+n*n,ab)-lower_bound(cd,cd+n*n,ab); } cout<<res<<endl; } int main() { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]); solve(); }
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