HDOJ 题目3908Triple（数学）

最新推荐文章于 2018-10-22 22:06:43 发布

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最新推荐文章于 2018-10-22 22:06:43 发布

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此博客讨论了如何计算给定整数集合中互质三元组的数量，包括求解方法和具体实现。

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Triple

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1303 Accepted Submission(s): 514

Problem Description

Given many different integers, find out the number of triples (a, b, c) which satisfy a, b, c are co-primed each other or are not co-primed each other. In a triple, (a, b, c) and (b, a, c) are considered as same triple.

Input

The first line contains a single integer T (T <= 15), indicating the number of test cases.
In each case, the first line contains one integer n (3 <= n <= 800), second line contains n different integers d (2 <= d < 10 ⁵) separated with space.

Output

For each test case, output an integer in one line, indicating the number of triples.

Sample Input

  
   1
6
2 3 5 7 11 13

Sample Output

Source

2011 Multi-University Training Contest 7 - Host by ECNU

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思路： http://blog.youkuaiyun.com/synapse7/article/details/20837477

ans=C(n,3) - 三个数中有一对互素的并且有一对不互素的个数sum

sum怎么算？

对于num[i]，计算与之互素的元素的个数pnum，那么不互素的元素的个数就是n-1-pnum了，对于num[i]而言，不满足条件的三元组个数为pnum*(n-1-pnum)。

注意最后还要除2：比如(2, 3, 4)在处理2的时候算了一次，在处理4的时候也算了一次，但是在处理3的时候，肯定不会出现，因为对3而言，2与3互素，4也与3互素，所以总的来说每个三元组都多算了一次。

#include<stdio.h>
#include<string.h>
int gcd(int a,int b)
{
	int t;
	if(a<b)
	{
		t=a;
		a=b;
		b=t;
	}
	if(b==0)
		return a;
	else
		return gcd(b,a%b);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,num,a[1000],i,j,sum=0,ans;
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n;i++)
		{
			num=0;
			for(j=0;j<n;j++)
			{
				if(i!=j)
					if(gcd(a[i],a[j])==1)
						num++;
			}
			sum+=num*(n-num-1);
		}
		ans=n*(n-1)*(n-2)/6-sum/2;
		printf("%d\n",ans);
	}
}