sgu 154 Factorial

154. Factorial

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

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

One number Q written in the input (0<=Q<=10^8).

Output

Write "No solution", if there is no such number N, and N otherwise.

Sample test(s)

Input

2

Output

10

问是否存在整数n，使得n！的末尾恰有Q个0.有的话输出最小满足条件的n

n！尾数0的个数就是 [n/5] + [n/25] + [n/5^3]+.....

我们可以二分枚举n ，然后做判断，不断缩小范围。。

知道AC。

贴个代码：

#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))#define FOR(a,b) for(int a=1;a<=(b);(a)++)using namespace std;int const nMax = 1010;int const base = 10;typedef long long LL;typedef pair<LL,LL> pij;//    std::ios::sync_with_stdio(false);long long n;LL f(LL a){    LL ans=0;    LL t=5;    while(a/t){        ans+=(a/t);        t*=5;    }    return ans;}int main(){    LL l,r;    l=0,r=450000000;    cin>>n;    LL ans=r;    while(l<=r){        LL mid=(l+r)>>1;        LL p=f(mid);       // printf("%lld %lld\n",mid,p);        if(p==n){            if(ans>mid)            ans=mid;            r=mid-1;        }else if(p>n){            r=mid-1;        }else {            l=mid+1;        }    }    if(ans==0)ans=1;    if(ans==450000000){        printf("No solution\n");    }else        printf("%I64d\n",ans);    return 0;}