light oj 1138 - Trailing Zeroes (III)(阶乘末尾0)

Posted 2020-10-19 风子磊

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	Output for Sample Input
3 1 2 5	Case 1: 5 Case 2: 10 Case 3: impossible

水题一发，直接二分查找。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
using namespace std;
LL n;
LL erfen()
{
    LL ans=-1, l=5, r=400000020;//想想r的值为什么这样定(因为此时r！等于10^8+1)
    while(l<=r)
    {
        LL mid=(l+r)/2;
        LL sum=0, s=mid;
        while(s>0)
            sum+=s/5,s/=5;
        if(sum==n)
        {
            r=mid-1;
            ans=mid;
        }
        else if(sum>n)
            r=mid-1;
        else
            l=mid+1;
    }
    return ans;
}
int main()
{
    int T, t=1;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld", &n);
        LL s=erfen();
        if(s!=-1)
        printf("Case %d: %lld\n", t++, s);
        else
            printf("Case %d: impossible\n", t++);
    }
}