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Description
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：给定n，求后导零个数为n的阶乘q!，输出q，不存在则输出“impossible";
思路：后导零，就是找2和5，由于5比2多所以直接找5，q!中有多少个因子5即有多少个后导零，每个5的倍数贡献一个后导零，每个25的倍数多贡献一个后导零，每个5^3又多贡献一个后导数零，以此类推.....阶乘q!后导零个数即为f(q)=q/5+q/25+q/125+.....。 然后二分解函数f(x)=n即可。
#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 namespace std; typedef long long ll; const int maxn=1000100; const ll INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); ll n; ll f(ll x) { ll res=0,t=5; while(t<=x){ res+=x/t; t*=5; } return res; } 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) return mid; else if(f(mid)<key) left=mid+1; else right=mid-1; } return -1; } int main() { int T,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); else puts("impossible"); } return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4567859.html
