lightoj 1138 （二分）

lightoj 1138

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

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】有些N！的末尾会是几个连续的0，找含Q个0的N！的最小N值；

【思路】用二分法，从区间【0， oo（无穷大）】查找，注意判断N！的末尾有几个0,。

AC代码：

#include<cstdio>#define LL long longLL sum(LL N)//求N阶乘中 末尾连续的0的个数{    LL ans = 0;    while(N)    {        ans += N / 5;        N /= 5;    }    return ans;}int k = 1;int main(){    int t;    LL Q;    scanf("%d", &t);    while(t--)    {        scanf("%lld", &Q);        LL left = 1, right = 1000000000000;        LL ans = 0;        while(right >= left)        {            int mid = (left + right) >> 1;            if(sum(mid) == Q)//相等时 要赋值给ans            {                ans = mid;                right = mid - 1;//找的是最小的ans            }            else if(sum(mid) > Q)                right = mid - 1;            else                left = mid + 1;        }        printf("Case %d: ", k++);        if(ans)            printf("%lld\n", ans);        else            printf("impossible\n");    }    return 0;}