Trailing Zeroes (III)————二分+尺取练习

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 N ! = 1 ∗ 2 ∗ . . . ∗ N .
For example, 5!=120, 5 ! = 120 , 120 contains one zero on the trail.

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

Each case contains an integer Q(1≤Q≤108) Q ( 1 ≤ Q ≤ 10 8 ) 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

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让你找阶乘末尾0个数是Q的数最小是多少，
不存在的话就输出impossible
关于阶乘末尾0的个数我这里有一篇博客讲怎么求：n！末尾0的个数

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具体看代码：

#include
using namespace std;
typedef long long ll;
const ll MAXN=10000000000000;

ll f(ll n)
{
    ll ans=0;
    while(n)    ans+=(n/=5);
    return ans;
}


int main()
{
    int t;ll Q;
    scanf("%d",&t);
    int kase=1;
    while(t--)
    {
        scanf("%lld",&Q);
        ll l=1,r=MAXN;
        ll ans=0;
        while(r>=l)
        {
            ll mid=(l+r)>>1;
            if(f(mid)==Q)
            {
                ans=mid;
//                break;//不能直接退出，因为阶乘末尾0个数等于Q的有很多呢，还得继续找最小的
                r=mid-1;//
            }
            else if(f(mid)1;
            else                r=mid-1;
        }
        printf("Case %d: ",kase++);
        if(ans) printf("%lld\n",ans);
        else    printf("impossible\n");
    }
    return 0;
}