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 = 1;
    while (m) {
        if (m & 1) ret = ret * n % mod;
        n = n * n % mod;
        m >>= 1;
    }
    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 = 0;
            rep2(i, 1, m)ans += quick_power(i, n, 1e9);
            int cnt = 0;
            while (ans) {
                if (ans % 10 != 0)break;
                cnt++;
                ans /= 10;
            }
            out << cnt << endl;
        }
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    TaskA solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}