hdu 5072 Coprime 容斥原理

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1509    Accepted Submission(s): 592

Problem Description

There are n people standing in a line. Each of them has a unique id number.

Now the Ragnarok is coming. We should choose 3 people to defend the evil. As a group, the 3 people should be able to communicate. They are able to communicate if and only if their id numbers are pairwise coprime or pairwise not coprime. In other words, if their id numbers are a, b, c, then they can communicate if and only if [(a, b) = (b, c) = (a, c) = 1] or [(a, b) ≠ 1 and (a, c) ≠ 1 and (b, c) ≠ 1], where (x, y) denotes the greatest common divisor of x and y.

We want to know how many 3-people-groups can be chosen from the n people.

 

 

Input

The first line contains an integer T (T ≤ 5), denoting the number of the test cases.

For each test case, the first line contains an integer n(3 ≤ n ≤ 10⁵), denoting the number of people. The next line contains n distinct integers a₁, a₂, . . . , a_n(1 ≤ a_i ≤ 10⁵) separated by a single space, where a_i stands for the id number of the i-th person.

 

 

Output

For each test case, output the answer in a line.

 

 

Sample Input

1

5

1 3 9 10 2

 

 

Sample Output

4

 

 

Source

2014 Asia AnShan Regional Contest

 

题目原形是同色三角形， 引用：就是求同色三角形的个数。总的三角形的个数是C(n,3)，只需减去不同色的三角形即可。对于每个点(数)，与它互质的连红边，不互质的连蓝边，那么对于该点不同色三角形个数为蓝边数*红边数/2，因为同一个三角形被计算了两次。那么同色三角形个数为C(n,3) - ∑蓝边数*红边数/2。

问题是：如何求 原来序列里面的n个数跟某个数k不互质的个数（互质的就是n-k了）？

可以将原来的n个数，每一个都把他们的不同的质因数都求出来，然后枚举它们能够组合的数（1 << cnt），用一个数组num记录，每枚举到一个数，那么数组对应的就+1

对于数k，也把它的不同质因数求出来，同样枚举它能够组合的所有数t，然后奇加偶减num

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int N = 200005;

int p[N][15], vis[N], a[N], num[N];
int n;
void Prime()
{
    memset(vis, 0, sizeof vis);
    for(int i = 0; i < N; ++i) p[i][0] = 0;
    for(int i = 2; i < N; ++i) if(!vis[i]) {
        p[i][ ++p[i][0] ] = i;
        for(int j = i + i; j < N; j += i) {
            vis[j] = 1;
            p[j][ ++p[j][0] ] = i;
        }
    }
    p[0][ ++p[0][0] ] = 1;    //考虑0的情况
}

void init()
{
    memset(num, 0, sizeof num);
    for(int k = 0; k < n; ++k)
    {
        int now = a[k];
        int cnt = p[ now ][0];
        for(int i = 1; i < (1 << cnt); ++i)
        {
            int t = 1;
            for(int j = 0; j < cnt; ++j) if((1 << j) & i) {
                t *= p[ now ][j + 1];
            }
            num[t]++;
        }
    }
}

void solve()
{
    ll ans = 0, res, sum = 0;
    ans = (ll)n * (n - 1) * (n - 2) / 6;    //类型转换一下，避免爆掉
    int tot = 0;
    for(int k = 0; k < n; ++k)
    {
        int now = a[k];
        int cnt = p[now][0];
        res = 0;
        for(int i = 1; i < (1 << cnt); ++i)
        {
            int t = 1, g = 0;
            for(int j = 0; j < cnt; ++j) if((1 << j) & i) {
                t *= p[ now ][j + 1];
                g++;
            }
            if(g & 1)  res += num[t];
            else       res -= num[t];
        }

        if(res == 0) continue;
        sum += (res - 1) * (n - res);
    }
    printf("%lld\n", ans  - sum / 2);

}
int main()
{
   // freopen("in", "r", stdin);
    int _;
    scanf("%d", &_);
    Prime();
    while(_ --)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
        init();
        solve();
    }
}