LIGHT_OJ 1138 the number of lead zeros after factorial

LIGHT_OJ 1138 the number of leading zeros after factorial n-trailing zeroes (III) Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%LLD &%llusubmit Status Practice LIGHTOJ 1138

Description

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

3

1

2

5

Sample Output

Case 1:5

Case 2:10

Case 3:impossible

Test instructions: Given n, the factorial q! of 0 number of n is obtained, the output Q, the output "impossible" is not present;

Train of thought: After guide zero, is looking for 2 and 5, because more than 5:2 so directly find 5,q! in how many factors 5 that is how many after guide zero, each 5 multiples contribute a post-zero, each of the multiples of 25 contribute a post-zero, each 5^3 contribute a post-derivative 0, and so on. Factorial q! after 0 number is f (q) =q/5+q/25+q/125+ ... Then the two decomposition function f (x) =n can be.

#include <iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<vector>#include<stack>#include<queue>#include<Set>#include<map>#include<string>#include<math.h>#include<cctype>using namespaceStd;typedefLong Longll;Const intmaxn=1000100;Constll inf= (1<< in);Const Doubleeps=0.0000000001;Const DoublePi=acos (-1.0); ll N;ll F (ll x) {ll res=0, t=5;  while(t<=x) {Res+=x/T; T*=5; }    returnRes;} ll Bin_search (ll left,ll right,ll key) { while(left<=Right ) {ll Mid= (left+right)/2; if(f (Mid) ==key&&f (mid-1) <key)returnmid; Else if(f (mid) <key) left=mid+1; Elseright=mid-1; }    return-1;}intMain () {intt,tag=1; CIN>>T;  while(t--) {cin>>N; ll ans=bin_search (1, Inf,n); printf ("Case %d:", tag++); if(ans!=-1) printf ("%lld\n", ans); ElsePuts"Impossible"); }    return 0;}

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LIGHT_OJ 1138 the number of lead zeros after factorial