LightOJ 1340 Story of Tomisu Ghost

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线性预处理出1~1e5的素数，nlogn的也可以

对于每组n，t，把n！质因数分解再贪心即可

还可以用下快速幂进行优化

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int MAXN = 1e5;const ll mod = 10000019;bool is_prime[MAXN + 5];vector<int> prime;void get_prime(){    memset(is_prime, true, sizeof(is_prime));    is_prime[0] = is_prime[1] = false;    for(int i = 2; i <= MAXN; i++)    {        if(is_prime[i]) prime.push_back(i);        for(int j = 0; j < prime.size() && i * prime[j] <= MAXN; j++)        {            is_prime[i * prime[j]] = false;            if(i % prime[j] == 0) break;        }    }}ll fast_pow(ll x, int n){    ll res = 1;    while(n)    {        if(n & 1) res = (res * x) % mod;        x = (x * x) % mod;        n >>= 1;    }    return res;}int cnt(int n, int p){    int res = 0;    while(n)        n /= p, res += n;    return res;}ll solve(int n, int t){    ll res = 1;    for(int i = 0; i < prime.size(); i++)    {        int p = prime[i], num = cnt(n, p);        if(num < t) break;        res = (res * fast_pow(p, num / t)) % mod;    }    if(res == 1) return -1;    return res;}int main(){    int T;    scanf("%d", &T);    get_prime();    for(int ca = 1; ca <= T; ca++)    {        int n, t;        scanf("%d%d", &n, &t);        printf("Case %d: %lld\n", ca, solve(n, t));    }    return 0;}

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