sgu 154

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=120,INF=0x7FFFFFFF;
lon arr[SZ];

lon getn(lon x)
{
    lon res=0;
    for(int i=1;i<=12;++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[1]=5;
        for(int i=2;i<=12;++i)arr[i]=arr[i-1]*5;
        lon lo=1,hi=1e12+10;
        for(;lo<hi;)
        {
            lon mid=(lo+hi)/2;
            if(getn(mid)<n)lo=mid+1;
            else hi=mid;
        }
        if(getn(lo)==n)cout<<lo<<endl;
        else cout<<"No solution"<<endl;
    }
    return 0;
}