NYOJ 1066 CO-PRIME（数论）

CO-PRIME

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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.

样例输入

31 2 3

样例输出

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)。

而根据上面的式子可以得出f(1)=∑mu[n] * F(n)。

所以把mu[]求出来，枚举n就行了，其中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;}