[UVA10139]Factovisors（数论，质因数）

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1080

题意：问n!能否被m整除。

给m分解质因数，用这些质因数以及他们的幂分别去除n，直到质因数的幂大于n。统计每一次的商，假如存在一个商小于质因数在m中出现的次数，说明n!不能被m整除了。uDEBUG好好用，没开LL一直wa，结果去查一下就查出来了。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
LL n, m;
vector<LL> p, a;

void f(LL x) {
  p.clear(); a.clear();
  LL xx = (LL)sqrt(x) + 1;
  for(LL i = 2; i < xx; i++) {
    if(x % i == 0) {
      p.push_back(i);
      LL cnt = 0;
      while(x % i == 0) {
        x /= i;
        cnt++;
      }
      a.push_back(cnt);
    }
  }
  if(x > 1) {
    p.push_back(x); a.push_back(1);
  }
}

LL mul(LL x, LL n) {
  LL ret = 1;
  while(n) {
    if(n & 1) ret *= x;
    n >>= 1; x *= x;
  }
  return ret;
}

int main() {
  //freopen("in", "r", stdin);
  //freopen("out", "w", stdout);
  while(~scanf("%lld%lld",&n,&m)) {
    if(m == 0) {
      printf("%lld does not divide %lld!
", m, n);
      continue;
    }
    f(m); bool flag = 0;
    for(LL i = 0; i < p.size(); i++) {
      LL cnt = 0;
      LL j = 1;
      LL q = mul(p[i], j++);
      while(n >= q) {
        cnt += (n / q);
        q = mul(p[i], j++);
      }
      if(cnt < a[i]) flag = 1;
      if(flag) break;
    }
    if(!flag) printf("%lld divides %lld!
", m, n);
    else printf("%lld does not divide %lld!
", m, n);
  }
  return 0;
}