4 Values whose Sum is 0
Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions: 32325 | Accepted: 9932 | |
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:有4个数列 每列找一个数,得到和为0的序列,有几种不同的方案
对1,2列的数求一个和,3,4列的数求一个和,然后进行二分查找
分析:如果一个一个枚举那么总共有n^4种情况 但是要全部判断一遍是不行的。不过将他们对半分成AB和CD在考虑,就可以快速地解决问题了。
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=4005;
int a[N],b[N],c[N],d[N],e[N*N];
int main()
{
int n;
while(scanf("%d",&n)==1){
for(int i=0;i<n;i++){
scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
e[i*n+j]=c[i]+d[j];
}
}
sort(e,e+n*n);
int cnt=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
int cd=-(a[i]+b[j]);
cnt+=upper_bound(e,e+n*n,cd)-lower_bound(e,e+n*n,cd);
}
}
printf("%d\n",cnt);
}
return 0;
}