【LightOJ】1138 - Trailing Zeroes (III)（数论，二分法）（POJ-1401类型题）

D - D

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

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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

这道题和poj的1401题很相似，额外再用了二分法，和高中二分法求根式的差不多。

代码如下：

#include <cstdio>#define MAX 0x7fffffff//最大的 long int型数 int cal(int n)//计算n的阶乘后0的个数{int ans = 0;while (n){ans += n/5;n /= 5;}return ans;}int main(){int u;int n;scanf ("%d",&u);int num = 1;while (u--){scanf ("%d",&n);printf ("Case %d: ",num++);int left = 0,right = MAX;int mid;while (left <= right){mid = (left + right)>>1;//就是(left + right)/2的意思if (cal(mid) >= n)right = mid - 1;elseleft = mid +1;}if (cal(left) == n)//若最后一次逼近的时候满足了条件，此时left==mid（无论left与right相差1还是相等） printf ("%d\n",left);elseprintf ("impossible\n");}return 0;}