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  • 4 Values whose Sum is 0(枚举+二分)

    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).

    思路:和之前更新的校赛的题目差不多,都是枚举+二分查找

    代码:

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    
    using namespace std;
    
    #define maxn 4040
    
    int a[maxn],b[maxn],c[maxn],d[maxn],sum[maxn*maxn];
    int main()
    {
    	int n,i,j;
    	while(scanf("%d",&n)!=EOF)
    	{
    		for(i=0;i<n;++i)
    		scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    		for(i=0;i<n;++i)
    		{
    			for(j=0;j<n;++j)
    				sum[n*i+j]=c[i]+d[j];
    		}
    		sort(sum,sum+n*n);
    		int ans=0;
    		for(i=0;i<n;++i)
    		{
    			for(j=0;j<n;++j)
    			{
    				int k=-(a[i]+b[j]);
    				ans+=upper_bound(sum,sum+n*n,k)-lower_bound(sum,sum+n*n,k);
    			} 
    		}
    		printf("%d
    ",ans);
    	}
    	return 0;
    } 
    
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781854.html
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