LightOJ 1138 Trailing Zeroes (III)

C - Trailing Zeroes (III)

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

Description

You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 10⁸) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

Sample Input

3

1

2

5

Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

题意：求末尾有Q个0，的N的阶乘，求N的最小值。

这里要用要用到一个定理

令F(x)表示正整数X末尾0个数，则有

当0<n<5时， f(n!)=0;

当N>=5时，f(n!)=k+f(k!),其中k=n/5;

#include<stdio.h>#include<string.h>#include<algorithm>#define LL long long using namespace std;LL sum(LL n){LL ans=0;while(n){ans+=n/5;n=n/5;}return ans;}int main(){LL p;int k=1;int t;scanf("%d",&t);while(t--){scanf("%lld",&p);LL left=1;LL right=1000000000000;LL ans=0;while(right>=left){LL mid=(right+left)/2;if(sum(mid)==p){ans=mid;right=mid-1;}else if(sum(mid)>p){right=mid-1;}elseleft=mid+1;}printf("Case %d: ",k++);if(ans)printf("%lld\n",ans);elseprintf("impossible\n");}return 0;}