4 Values whose Sum is 0
Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions:28912 | Accepted: 8745 | |
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).
折半。先考虑 a+b 在考虑c+d
二分查找第二个中是否有满足和为0的数。
利用了 STL 中的 upper_bound lower_bound
STL中的函数:(使用前要先排序。)
1.lower_bound算法返回一个非递减序列[first, last)中的第一个大于等于val的位置,
2.upper_bound算法返回一个非递减序列[first, last)中的第一个大于val的位置。
如果所有元素都小于val, 则返回last的位置;
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <queue>
#include <cmath>
using namespace std;
#define N 4500
#define ll long long
int a[N],b[N],c[N],d[N];
int num[N*N],n;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int cnt=0,ans=0;
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++)
{
num[cnt++]=a[i]+b[j];
}
}
sort(num,num+cnt);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
int temp=c[i]+d[j];
ans+=(upper_bound(num,num+cnt,-temp)-lower_bound(num,num+cnt,-temp));
}
}cout<<ans<<endl;
return 0;
}