Codeforces Round #450 (Div. 2) D. Unusual Sequences 莫比乌斯系数容斥

D. Unusual Sequences
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Count the number of distinct sequences a1, a2, …, an (1 ≤ ai) consisting of positive integers such that gcd(a1, a2, …, an) = x and . As this number could be large, print the answer modulo 109 + 7.

gcd here means the greatest common divisor.

Input
The only line contains two positive integers x and y (1 ≤ x, y ≤ 109).

Output
Print the number of such sequences modulo 109 + 7.

Examples
input
3 9
output
3
input
5 8
output
0
题意：构造一个数组，要求每一位大于等于1，所有数的gcd为1，就相当于构造y/x,gcd为1的序列，首先我们可以证明对于n，把它任意的分成一个数组，一共有2^(n-1)种，然后可以知道这些序列的gcd是某一个数a的倍数就是2^(n/a-1)种，然后用莫比乌斯系数容斥就可以得到答案了。因为对于1e9的数据来说，最多只有11个质因数，那么可以用状压处理处需要的莫比乌斯系数的值，然后可以得到答案。

#include<bits/stdc++.h>#define ll long longusing namespace std;vector<int> pe;vector<int> pr;bool vis[100005];const int mod = 1e9+7;int x,y;ll qpow(int x,int b){    ll now = x;    ll sum =1;    while(b){        if(b&1) sum = sum*now%mod;        now = now*now%mod;        b >>= 1;    }    return sum;}void init(){    vis[1] = true;    for(int i = 2;i <= 100005;i ++){        if(vis[i] == false) pe.push_back(i);        for(int j = 0;j < pe.size()&&i*pe[j] < 100005;j ++){            vis[i*pe[j]] = true;            if(i%pe[j] == 0) break;        }    }    int now = x;    for(int i =0;i < pe.size();i ++){        if(now%pe[i] == 0){            pr.push_back(pe[i]);            while(now%pe[i]==0) now /= pe[i];        }    }    if(now != 1) pr.push_back(now);    int res = pr.size();    ll ans = qpow(2,x-1);    for(int i = 1;i < (1<<res);i ++){        int tmp = 1,f = 1;        for(int j = 0;j <res;j ++){            if(i&(1<<j)) tmp *= pr[j],f *= -1;        }        ans = (ans+qpow(2,x/tmp-1)*f+mod)%mod;    }    printf("%lld\n",ans);}int main(){    scanf("%d %d",&x,&y);    if(y%x){        puts("0");        return 0;    }    x = y/x;    init();    return 0;}