(数学题) Counting

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概述

问题描述:

描述

Given many 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.

输入

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 <= 1000), second line contains n integers d (1 <= d < 105) separated with space.

输出

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

样例输入

1
6
2 3 5 7 11 13

样例输出

思路. 本题意在找出所有的三元组,满足:

1 三个数都互质

2 三个数都不互质

换一种思路:

当三元组中存在一个或两个数对不互质时,不满足条件 .

假设a 为与元素i互质的元素个数，b为与元素i不互质的元素个数，那么包含元素i 的，不满足条件的三元组共a*b个。

需通过观察可知，当一次求完a*b累加后，正好多算了一倍。也就是结果应为：总情况数 - sum(a*b)/2;

代码:

#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#include <utility>
#include <cstring>
using namespace std;
int gcd (int v1,int v2)
{
    while(v2){
        int tmp = v2;
        v2 = v1 % v2;
        v1 = tmp;
    }
    return v1;
}

int main(int argc, char* argv[])
{
    int t;
    cin >> t;
    while(t--)
    {
        vector<int> num;
        int n;
        cin >> n;
        for(int i=0;i<n;i++)
        {
            int tmp;
            scanf("%d",&tmp);
            num.push_back(tmp);
        }
        int cgcd=0,cngcd=0,total=(n*(n-1)*(n-2)/6),ans=0;;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)
            if(i!=j){
                if(gcd(num[i],num[j])==1)
                    cgcd++;
                else
                cngcd++;}
            ans+=cgcd*cngcd;
            cgcd=0;cngcd=0;
        }
        cout<< total-ans/2<<endl;
    }
}