POJ - 2785 :4 Values whose Sum is 0 (二分)
阿新 • • 發佈:2017-07-26
num pro ive con res per cin 查找 ati ) that belong respectively to A, B, C and D .
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 28Output
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). 輸入四組數,問在四組數每組取一個數相加等於0的方法數。 將前兩組和枚舉排序,後兩組和枚舉。 二分查找排序後的數組即可,復雜度O(N^2) 源代碼:#include<iostream>
#include<algorithm>
using namespace std;
int a[4005],b[4005],c[4005],d[4005],e[16000005],f[16000005];
int main()
{
int n;
while(cin>>n)
{
int x=0,y=0,z,t;
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++)
{
e[x++]=a[i]+b[j];
f[y++]=c[i]+d[j];
}
sort(e,e+x);
long long int s=0;
for(int i=0;i<y;i++)
{
s+=upper_bound(e,e+n*n,-f[i])-lower_bound(e,e+n*n,-f[i]);
}
cout<<s<<endl;
}
return 0;
}
POJ - 2785 :4 Values whose Sum is 0 (二分)