CF-1114C-Trailing Loves (or L'oeufs?)

题意：

输入n和m，求n!转换成m进制之后末尾有多少个0；

思路：

转换一下题意就可以看成，将n表示成x * (m ^ y)，求y的最大值。^表示次方而不是异或；

这就比较好想了，将m分解质因数，对于每个质因数，设n!含有a个，m含有b个，则ans = min(ans, a / b);

• 自己比赛的时候写的

  C - Trailing Loves (or L'oeufs?)

  GNU C++11

  Accepted

  46 ms

  0 KB

  #include "bits/stdc++.h"
  using namespace std;
  typedef long long LL;
  map<LL, LL> mp, num;
  vector<LL> vc;
  int main() {
      LL n, m;
      scanf("%lld%lld", &n, &m);
      for (LL i = 2; i * i <= m; i++) {
          if (m % i == 0) {
              vc.push_back(i);
              while (m % i == 0) {
                  m /= i;
                  mp[i]++;
              }
          }
      }
      if (m != 1) {
          vc.push_back(m);
          mp[m]++;
      }
      for (int i = 0; i < vc.size(); i++) {
          LL j = 1;
          while (j <= n / vc[i]) {
              j *= vc[i];
              num[vc[i]] += n / j;
          }
      }
      LL ans = 1LL << 62;
      for (LL i = 0; i < vc.size(); i++) {
          ans = min(ans, num[vc[i]] / mp[vc[i]]);
      }
      printf("%lld\n", ans);
      return 0;
  }

  其实没必要把各个因子保存下来。标程还是优很多的

• 看了标程之后改的

  C - Trailing Loves (or L'oeufs?)

  GNU C++11

  Accepted

  31 ms

  0 KB

  #include "bits/stdc++.h"
  using namespace std;
  typedef long long LL;
  const LL INF = 1LL << 60;
  int main() {
      LL n, m, ans = INF;
      scanf("%lld%lld", &n, &m);
      for (LL i = 2; i <= m; i++) {
          if (i * i > m) {
              i = m;
          }
          if (m % i == 0) {
              int cnt = 0;
              while (m % i == 0) {
                  m /= i;
                  cnt++;
              }
              LL tmp = 0, mul = 1;
              /*
               for (LL mul = i; mul <= n; mul *= i)
               这种写法应该更符合正常思维，但是因为n最高可以达到1e18，比较接近LL上限，mul可能乘i之前还小于n,乘完就爆LL了； 
              */ 
              while (mul <= n / i) {
                  mul *= i;
                  tmp += n / mul;
              }
              ans = min(ans, tmp / cnt);
          }
      }
      printf("%lld\n", ans);
      return 0;
  }