??? cliquers

解：先推一个式子，然后就是CRT了...

那个阶乘怎么求呢？主要是分母可能有0，这时我们把分母的因子p全部提出来，上下次数相减判断即可。

细节颇多......注意在快速幂开始的时候a %= MO是个好习惯。

#include <cstdio>
#include <algorithm>

typedef long long LL;
const int N = 100010;
const LL MO = 1e9 - 401, mod[] = {0, 2, 13, 5281, 7283};

int turn;
LL ans[5], n, m, nn[10000][5];
//LL inv[10000][5], invn[10000][5];

inline LL qpow(LL a, LL b, LL c) {
    LL ans = 1;
    a %= c;
    while(b) {
        if(b & 1) ans = ans * a % c;
        a = a * a % c;
        b = b >> 1;
    }
    return ans;
}

inline LL Pow(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if(b & 1) ans = ans * a;
        a = a * a;
        b = b >> 1;
    }
    return ans;
}

LL exgcd(LL a, LL &x, LL b, LL &y) {
    if(!b) {
        x = 1; y = 0;
        return a;
    }
    LL g = exgcd(b, x, a % b, y);
    std::swap(x, y);
    y -= x * (a / b);
    return g;
}

inline LL getnn(LL x) {
    if(!x) return 1;
    LL t = x / mod[turn];
    LL ans = qpow(nn[mod[turn] - 1][turn], t, mod[turn]);
    t = x % mod[turn];
    return ans * nn[t][turn] % mod[turn] * getnn(x / mod[turn]);
}

inline LL cal(LL k) {

    //printf("cal %lld 
", k);
    //bool f = (turn == 2 && k == 2) || (turn == 2 && k == 1);

    LL time = 0;
    for(LL i = n; i > 1; i /= mod[turn]) {
        time += i / mod[turn];
        //printf("1 time += %lld  = %lld 
", i / mod[turn], time);
    }
    for(LL i = k; i > 1; i /= mod[turn]) {
        time -= i / mod[turn];
        //printf("2 time -= %lld  = %lld 
", i / mod[turn], time);
    }
    for(LL i = n / k; i > 1; i /= mod[turn]) {
        time -= (i / mod[turn]) * k;
        //printf("3 time -= %lld  = %lld
", i / mod[turn], time);
    }
    if(time) {
        //printf(" -- -- return 0 
");
        //printf("mod = %lld ans = 0
", mod[turn]);
        return 0;
    }

    LL t1 = getnn(n), t2 = getnn(k), t3 = getnn(n / k);
    /*if(f) {
        printf("%lld %lld %lld 
", t1, t2, t3);
    }*/
    t3 = qpow(t3, k, mod[turn]);
    t2 = qpow(t2, mod[turn] - 2, mod[turn]);
    t3 = qpow(t3, mod[turn] - 2, mod[turn]);

    //printf("return %lld 
", t1 * t2 % mod[turn] * t3 % mod[turn]);
    //printf("mod = %lld k = %lld ans = %lld 
", mod[turn], k, t1 * t2 % mod[turn] * t3 % mod[turn]);
    return t1 * t2 % mod[turn] * t3 % mod[turn];
}

inline LL solve() { /// cal each prime
    LL ans = 0;
    for(LL i = 1; i * i <= n; i++) {
        if(n % i) continue;
        ans = (ans + cal(i)) % mod[turn];
        if(i * i < n) {
            ans = (ans + cal(n / i)) % mod[turn];
        }
    }
    return ans;
}

inline void work() {
    scanf("%lld%lld", &n, &m);
    //m %= MO;

    for(turn = 1; turn <= 4; turn++) {
        ans[turn] = solve();
        //printf("ans %d %lld = %lld 
", turn, mod[turn], ans[turn]);
    }


    /// excrt
    /*LL a = ans[1], p = mod[1];
    for(int i = 2; i <= 4; i++) {
        /// merge
        //printf("merge %d 
", i);
        LL P = p * mod[i], x, y;
        LL c1 = ((a - ans[i]) % P + P) % P;
        //printf("a = %lld  c1 = %lld 
", a, c1);
        LL g = exgcd(mod[i], x, p, y);
        (x *= (c1 / g)) %= P; (y *= (c1 / g)) %= P;
        a = (x * mod[i] % P + ans[i]) % P;
        p = P;
        //printf("merge %lld ans = %lld 
", mod[i], a);
    }*/
    /// crt
    LL a = 0, p = MO - 1;
    for(int i = 1; i <= 4; i++) {
        a = (a + ans[i] * (p / mod[i]) % p * qpow(p / mod[i], mod[i] - 2, mod[i]) % p) % p;
    }

    //printf("over 
");
    a = (a % p + p) % p;
    //printf("a = %lld
", a);
    LL ans = qpow(m, a, MO);
    printf("%lld
", ans);
    return;
}

inline void prework() {
    for(turn = 1; turn <= 4; turn++) {
        //invn[0][turn] = inv[0][turn]turn] = nn[0][turn] = 1;
        nn[0][turn] = 1;
        //invn[1][turn] = inv[1][turn]turn] = nn[1][turn] = 1;
        for(int i = 1; i < mod[turn]; i++) {
            nn[i][turn] = nn[i - 1][turn] * i % mod[turn];
            //inv[i][turn] = inv[mod[turn] % i] * (mod[turn] - mod[turn] / i) % mod[turn];
            //invn[i][turn] = invn[i - 1][turn] * inv[i][turn] % mod[turn];
        }
    }
    return;
}

int main() {

    freopen("cliquers.in", "r", stdin);
    freopen("cliquers.out", "w", stdout);

    prework();

    int T;
    scanf("%d", &T);
    while(T--) {
        work();
    }
    return 0;
}

两种CRT都写了。