标签：tput nat std exactly == min lld pre ...
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：给出数字，代表某个数的阶乘末尾连续0的个数，求出这个数是多少
代码：
1 #include<stdio.h> 2 int num(int n) //求n的阶乘末尾连续0的个数 3 { //百度说是定理 记住吧... 4 5 int ans=0; 6 while(n) 7 {ans+=n/5; 8 n=n/5; 9 10 } 11 return ans; 12 13 } 14 int main() 15 { 16 int mid,i=1; 17 int t ,q; 18 long long l,r; 19 scanf("%d",&t); 20 while(t--) 21 {scanf("%d",&q); 22 l=0; 23 r=100000000000000; //r要大于1e8 24 long long m=0; 25 while(l<=r) 26 {mid=(l+r)/2; 27 if(num(mid)==q) 28 {r=mid-1; 29 m=mid; 30 } 31 else 32 {if(num(mid)>q) 33 r=mid-1; 34 else l=mid+1; 35 } 36 37 } 38 39 if(m==0) printf("Case %d: impossible\n",i); 40 else printf("Case %d: %lld\n",i,m); 41 i++; 42 } 43 return 0; 44 }
标签：tput nat std exactly == min lld pre ...
原文地址：http://www.cnblogs.com/hss-521/p/7255335.html
