题意：

询问有多少数\(n\)满足\(n^{n!}\equiv b\mod p \land\ n\in[1,M]\)，数据范围：\(M\leq2^{64}-1,p\leq1e5\)

思路：

这题显然要用欧拉降幂，\(n!\)小于\(\varphi(p)\)的直接暴力算，\(n!\neq 0\mod \varphi(p)\)也直接暴力。

\(n!\equiv 0\mod \varphi(p)\)显然这时质数恒为\(\varphi(p)\)，由鸽笼定理得：

当\(x\)是常数时，\(1^x,2^x,\dots,n^x,\dots\mod p\)有循环节为\(\varphi(p)\)

那么直接按循环节搞一下即可。

注意一下，当\(b=0,M=2^{64}-1,p=1\)时，答案爆\(long long\)。

代码：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e5 + 5;
const int MAXM = 3e6;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
ull euler(ull n){
    ull res = n, a = n;
    for(int i = 2; i * i <= a; i++){
        if(a % i == 0){
            res = res / i * (i - 1);
            while(a % i == 0) a/= i;
        }
    }
    if(a > 1) res = res / a * (a - 1);
    return res;
}
ull ppow(ull a, ull b, ull mod){
    ull ret = 1;
    while(b){
        if(b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
ull rec[maxn];
int main(){
    int T, ca = 1;
    scanf("%d", &T);
    while(T--){
        ull b, p, m;
        scanf("%I64u%I64u%I64u", &b, &p, &m);
        if(b == 0 && p == 1){
            if(m == 18446744073709551615ULL)
                printf("Case #%d: 18446744073709551616\n", ca++);
            else
                printf("Case #%d: %I64u\n", ca++, m + 1);
            continue;
        }
        ull phi = euler(p);
        ull ans = 0, fac = 1;
        ull i = 1;
        if(b == 0) ans++;
        for(i = 1; i <= m; i++){
            if(fac * i >= phi) break;
            fac = fac * i;
            if(ppow(i, fac, p) == b) ans++;
        }
        for(; i <= m; i++){
            if(fac * i % phi == 0) break;
            fac = fac * i % phi;
            if(ppow(i, fac + phi, p) == b) ans++;
        }
        if(i <= m){
            ull cnt = 0;
            for(int j = 1; j <= p; j++){
                rec[j] = ppow(j, phi, p);
                if(rec[j] == b) cnt++;
            }
            for(; i <= p && i <= m; i++){
                if(rec[i] == b) ans++;
            }
            if(i <= m){
                ull rest = m - p;
                ans += rest / p * cnt;
                rest -= rest / p * p;
                for(i = 1; i <= rest; i++){
                    if(rec[i] == b) ans++;
                }
            }
        }
        printf("Case #%d: %I64u\n", ca++, ans);
    }
    return 0;
}