快速幂

int fastPow(int a, int x, int p) {
    if (x==0) {
        return 1%p;
    }
    int tmp=fastPow(a,x/2,p);
    tmp=(long long)tmp*tmp%p;
    if (x%2==1) {
        tmp=(long long)tmp*a%p;
    }
    return tmp;
}

int fastPow2(int a, int x, int p) {
    int s=1%p;
    for (;x;x>>=1) {
        if (x&1) {
            s=(long long)s*a%p;
        }
        a=(long long)a*a%p;
    }
    return s;
}