ZOJ3549 Little Keng(快速幂)

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Little Keng

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Calculate how many 0s at the end of the value below:
1ⁿ + 2ⁿ + 3ⁿ + ... + mⁿ

Input

There are multiple cases. For each cases, the input containing two integers m, n. (1 <= m <= 100 , 1 <= n <= 1000000)

Output

One line containing one integer indicatint the number of 0s.

Sample Input

4 3

Sample Output

2

看完这题，感觉0的位数不会很多，拿Java大数跑了一部分数据，发现就直接取1e9没什么问题，然后就直接搞了。

 /**
  * code generated by JHelper
  * More info: https://github.com/AlexeyDmitriev/JHelper
  * @author xyiyy @https://github.com/xyiyy
  */

 #include <iostream>
 #include <fstream>

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>

 using namespace std;
 #define rep2(X, L, R) for(int X=L;X<=R;X++)
 typedef long long ll;

 //
 // Created by xyiyy on 2015/8/5.
 //

 #ifndef ICPC_QUICK_POWER_HPP
 #define ICPC_QUICK_POWER_HPP
 typedef long long ll;

 ll quick_power(ll n, ll m, ll mod) {
     ll ret = ;
     while (m) {
         if (m & ) ret = ret * n % mod;
         n = n * n % mod;
         m >>= ;
     }
     return ret;
 }

 #endif //ICPC_QUICK_POWER_HPP

 class TaskA {
 public:
     void solve(std::istream &in, std::ostream &out) {
         int m, n;
         while (in >> m >> n) {
             ll ans = ;
             rep2(i, , m)ans += quick_power(i, n, 1e9);
             int cnt = ;
             while (ans) {
                 if (ans %  != )break;
                 cnt++;
                 ans /= ;
             }
             out << cnt << endl;
         }
     }
 };

 int main() {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     TaskA solver;
     std::istream &in(std::cin);
     std::ostream &out(std::cout);
     solver.solve(in, out);
     return ;
 }