luoguP2231 [HNOI2002]跳蚤

题目链接

bzoj1220: [HNOI2002]跳蚤

题解

根据裴蜀定理，不定方程的解为未知数的gcd，所以选取的n个数的gcd为1

那么n - 1个数保证没有公约数为m的约数，枚举质因数容斥

质因数的个数上届是log的啊，我真傻，还想了半天QAq

那啥，bzoj高精，你们去做吧Qwq

代码

#include<cstdio>
#include<algorithm>
#define LL long long
inline LL read() {
    LL x = 0,f = 1;
    char c = getchar();
    while(c < '0' || c > '9')c = getchar();
    while(c <= '9' && c >= '0')x = x * 10 + c - '0',c = getchar();
    return x * f;
}
const int maxn = 100007;
LL ans = 0,n,m;
LL pow(LL x,LL res) {
    LL ret =  1;
    for(;res;res >>= 1,x *= x) if(res & 1) ret *= x ;
    return ret;
}
int num = 0,rime[maxn];
void divede(int x) {
    for(int i = 2;i * i <= x;++ i)
        if(x % i == 0) {
            rime[++ num] = i;
            while(x % i == 0) x /= i;
        }
    if(x != 1) rime[++ num]  = x;
}
void solve(int cnt = 1,LL x = 1,int tp = 1) {
    if(cnt == num + 1) { ans += (tp & 1) ? pow(m / x,n) : -pow(m / x,n); return ;}
    solve(cnt + 1,x,tp);
    solve(cnt + 1,x * rime[cnt],tp ^ 1);
}
int main() {
    n = read(),m = read();
    //ans = 2 * pow(m,n);
    divede(m);
    solve();
    printf("%lld\n",ans);
    return 0;
}