二分-G - 4 Values whose Sum is 0

G - 4 Values whose Sum is 0

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 ²⁸ ) 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).

　　题目大意：每行会给出四个整数，要求在每列整数中找出一个整数，使四个整数之和为0，求有几种不同组合

 #include<iostream>

 #include<algorithm>

 using namespace std;

 const int maxn = ;

 const int maxm = *;

 int a[maxn],b[maxn],c[maxn],d[maxn];

 int ab[maxm],cd[maxm];

 int main(){

     int n;

     scanf("%d",&n);

     for(int i=;i<n;i++)

         scanf("%d %d %d %d",a+i,b+i,c+i,d+i);

     int cnt = ;

     for(int i=;i<n;i++)//前两列整数求和

         for(int j=;j<n;j++)

             ab[cnt++] = a[i] + b[j];

     cnt = ;

     for(int i=;i<n;i++)//后两列整数求和

         for(int j=;j<n;j++)

             cd[cnt++] = c[i] + d[j];

     sort(ab,ab+n*n);

     sort(cd,cd+n*n);

     int p = n*n-;

     cnt = ;

     for(int i=;i<n*n;i++){

         for(;p>=;p--)

             if(ab[i]+cd[p]<=)    break;

         if(p<)                    break;

         for(int j=p;j>=;j--){

             if(ab[i]+cd[j]==)    cnt++;

             if(ab[i]+cd[j]<)    break;

         }

     }

     printf("%d",cnt);

 }

　　面对四列整数，一一组合将会造成极其庞大的数据，可将其中两两组合，组合以后再相加寻找和为0的情况。

巴特西

二分-G - 4 Values whose Sum is 0

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