Lightoj Trailing Zeroes (III) 1138 "binary search + factorial decomposition"

1138-trailing Zeroes (III)

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Time Limit: 2 second (s)

Memory Limit: MB

You task was 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! = 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

problem Setter:jane ALAM JAN

Test instructions: Give you a number, this number represents n! There are several 0 behind. Give this number and calculate the value of N.

Problem Solving Ideas:

Any factorization can be written in the form of prime number multiplication. So calculate the factorial of a number after a few 0, just calculate how much this number contains 5. (The point about this is unclear: click the Open link).

You can use the dichotomy to find this point. The idea of using two points for this problem is no more difficult.

AC Code;

<span style= "FONT-SIZE:18PX;" > #include <stdio.h> #include <math.h> #include <vector> #include <queue> #include <string > #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm>using Namespace Std;typedef long long LL;    ll solve (ll N) {ll num=0;        while (n) {NUM+=N/5;    n/=5; } return num;}    ll ER (ll N) {ll x=1;    LL y=1844674407370;    LL mid;    LL Res=-1;        while (y>=x) {mid= (x+y)/2;        LL Ans=solve (mid);            if (ans==n) {res=mid;            Y=mid-1;        return mid;        } else if (ans>n) {y=mid-1;        } else if (ans<n) {x=mid+1; }} return res;}    int main () {int t;    scanf ("%d", &t);    int xp=1;        while (t--) {LL n;        scanf ("%lld", &n);        LL ans=er (n);        if (ans==-1) printf ("Case%d:impossible\n", xp++);    else printf ("Case%d:%d\n", Xp++,ans); } return0;} </span>

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Lightoj Trailing Zeroes (III) 1138 "binary search + factorial decomposition"