NYOJ 1066 CO-PRIME（数论）

CO-PRIME

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 3

描写叙述

This problem is so easy! Can you solve it?

You are given a sequence which contains n integers a1,a2……an, your task is to find how many pair(ai, aj)(i < j) that ai and aj is co-prime.

 

输入

There are multiple test cases.
Each test case conatains two line,the first line contains a single integer n,the second line contains n integers.
All the integer is not greater than 10^5.

输出

For each test case, you should output one line that contains the answer.

例子输入

3
1 2 3

例子输出

3

题意：给出n个正整数，求这n个数中有多少对互素的数。

分析：莫比乌斯反演。

此题中，设F(d)表示n个数中gcd为d的倍数的数有多少对，f(d)表示n个数中gcd恰好为d的数有多少对，

则F(d)=∑f(n) (n % d == 0)

  f(d)=∑mu[n / d] * F(n) (n %d == 0)

上面两个式子是莫比乌斯反演中的式子。

所以要求互素的数有多少对。就是求f(1)。

而依据上面的式子能够得出f(1)=∑mu[n] * F(n)。

所以把mu[]求出来，枚举n即可了。当中mu[i]为i的莫比乌斯函数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN = 1e5 + 10;
typedef long long LL;
int cnt[MAXN], pri[MAXN], num[MAXN], pri_num, mu[MAXN], vis[MAXN], a[MAXN];

void mobius(int n)  //筛法求莫比乌斯函数
{
    pri_num = 0;
    memset(vis, 0, sizeof(vis));
    vis[1] = mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            pri[pri_num++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < pri_num; j++) {
            if(i * pri[j] > n) break;
            vis[i*pri[j]] = 1;
            if(i % pri[j] == 0) {
                mu[i*pri[j]] = 0;
                break;
            }
            mu[i*pri[j]] = -mu[i];
        }
    }
}

LL get(int x)
{
    return (LL)x * (x-1) / 2;
}

int main()
{
    mobius(100005);
    int n;
    while(~scanf("%d",&n)) {
        int mmax = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d",&a[i]);
            mmax = max(mmax, a[i]);
        }
        memset(cnt, 0, sizeof(cnt));
        memset(num, 0, sizeof(num));
        for(int i = 1; i <= n; i++) num[a[i]]++;
        for(int i = 1; i <= mmax; i++)
            for(int j = i; j <= mmax; j += i)
                cnt[i] += num[j];
        LL ans = 0;
        for(int i = 1; i <= mmax; i++)
            ans += get(cnt[i]) * mu[i];
        printf("%lld\n", ans);
    }
    return 0;
}