ZOJ 4528 ZOJ Monthly, October 2011 A

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

这套题很多数论啊。

这个题目直接暴力即可~

原题等价于那个数列的和能够被10^k整除的最大k是多少，枚举就可以了。

但是要用大数。

我的代码：

import java.math.*; import java.util.*; public class Main { static BigInteger power(BigInteger p,BigInteger n,BigInteger m) { BigInteger sq=BigInteger.ONE; while(n.compareTo(BigInteger.ZERO)>0) { if(n.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ONE)==0) sq=sq.multiply(p).mod(m); p=p.multiply(p).mod(m); n=n.divide(BigInteger.valueOf(2)); } return sq.mod(m); } static boolean judge(int m,int n,int k) { BigInteger mod=BigInteger.ONE,ans=BigInteger.ZERO; int i; for(i=1;i<=k;i++) mod=mod.multiply(BigInteger.valueOf(10)); for(i=1;i<=m;i++) { BigInteger a,b; a=BigInteger.valueOf(i); b=BigInteger.valueOf(n); ans=(ans.add(power(a,b,mod))).mod(mod); } if(ans.mod(mod).compareTo(BigInteger.ZERO)==0) return true; else return false; } public static void main(String args[]) { Scanner cin=new Scanner(System.in); int i,m,n; while(cin.hasNext()) { m=cin.nextInt(); n=cin.nextInt(); for(i=1;;i++) { if(judge(m,n,i)) continue; else break; } System.out.println(i-1); } } }