POJ2785 4 Values whose Sum is 0(暴力二分查找)

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• 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).

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  思路：

  题目是要求四个数相加之和为零， 我们可以两 两相加， 这样的话就变成是求两个数之和为0了，再进一步想就是在另一个数组中找相反数，利用low_bound

  ()和up_bound可以快速的二分查找。

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  #include<iostream>
  #include<cstdio>
  #include<algorithm>
  
  using namespace std;
  const int mx = 4050;
  int a[mx], b[mx], c[mx], d[mx];
  int n, sum1[mx * mx], sum2[ mx * mx];
   void slove(){
   	 int cnt =0, nn = n * n;
   	 sort(sum1, sum1 + nn);
   	 
  	 for(int i = 0; i < nn; i++){
  	 	cnt += upper_bound(sum1, sum1+ nn, -sum2[i]) - lower_bound(sum1, sum1+ nn, -sum2[i]); //记得负号 
  	 }
  	 cout<<cnt<<endl;
   }
  int main(){
  	while(scanf("%d", &n) != EOF){
  		
  		for(int i = 0; i < n; i++)
  			scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
  	
  		int c1 = 0, c2 = 0; 
  		for(int i =0; i < n; i++)
  			for(int j =0; j < n; j++){
  			sum1[c1++] = a[i] + b[j];
  			sum2[c2++] = c[i] + d[j];
  		}	
  		
  		slove();
  	}
  
  	return 0;
  }
  

  
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