题意
题目链接
Sol
首先若y % x不为0则答案为0
否则，问题可以转化为，有多少个数列满足和为y/x，且整个序列的gcd=1
考虑容斥，设(g[i])表示满足和为(i)的序列的方案数，显然(g[i] = 2^{i-1})(插板后每空位放不放)
同时还可以枚举一下gcd，设(f[i])表示满足和为(i)且所有数的gcd为1的方案，(g[i] = sum_{d | i} f[frac{n}{d}])
反演一下，(f[i] = sum_{d | i} mu(d) g(frac{i}{d}))
mu函数可以暴力枚举质因子得到
复杂度(O(2^{Mx} * Mx sqrt{N}) ，(Mx)最大为10
代码语言：javascript复制#include<bits/stdc .h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1 )
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e6 10, mod = 1e9 7, INF = 1e9 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x y < 0) return x y mod; return x y >= mod ? x y - mod : x y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x y < 0) x = x y mod; else x = (x y >= mod ? x y - mod : x y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod mod) % mod;}
template <typename A> inline void debug(A a){cout << a << 'n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 c - '0', c = getchar();
return x * f;
}
int x = read(), y = read();
map<int, int> mu;
int g(int a, int p) {
int base = 1;
while(p) {
if(p & 1) base = mul(base, a);
p >>= 1; a = mul(a, a);
}
return base;
}
signed main() {
if(y % x != 0) return puts("0"), 0;
vector<int> d; y /= x; int p = y;
for(int i = 2; i * i <= y; i )
if(!(y % i)) {
d.push_back(i);
while(!(y % i)) y /= i;
}
if(y != 1) d.push_back(y);
y = p;
for(int sta = 0; sta < (1 << d.size()); sta ) {
int v = 1, t = 1;
for(int i = 0; i < d.size(); i ) if(sta & (1 << i)) t *= -1, v *= d[i];
mu[v] = t;
}
int ans = 0;
for(auto &x: mu) {
int d = x.fi, m = x.se;
add2(ans, mul(m mod, g(2, y / d - 1)));
}
cout << ans;
return 0;
}
