Factorial
题意：能否找到一个数，它的阶乘后面0的个数为n?
数越大，阶乘后的0越多。用二分找。对于一个数x，它的阶乘，将小于等于它的数分解质因数。其中2的个数一定大于5的个数。因此计5的个数就是结果末尾0的个数。比它小的数有x/5个5的倍数，x/25个25的倍数。那么5的数量就是x/5+x/25......
//#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include <stack>
using namespace std;
typedef long long lon;
const lon SZ=,INF=0x7FFFFFFF;
lon arr[SZ]; lon getn(lon x)
{
lon res=;
for(int i=;i<=;++i)
{
res+=x/arr[i];
}
return res;
} int main()
{
//std::ios::sync_with_stdio(0);
//freopen("d:\\1.txt","r",stdin);
//for(;scanf("%d",&n)!=EOF;)
lon casenum;
//cin>>casenum;
//for(lon time=1;time<=casenum;++time)
{
lon n;
cin>>n;
arr[]=;
for(int i=;i<=;++i)arr[i]=arr[i-]*;
lon lo=,hi=1e12+;
for(;lo<hi;)
{
lon mid=(lo+hi)/;
if(getn(mid)<n)lo=mid+;
else hi=mid;
}
if(getn(lo)==n)cout<<lo<<endl;
else cout<<"No solution"<<endl;
}
return ;
}