154. Factorial （二分+数论）

154. Factorial

time limit per test: 0.25 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

 

//首先在阶乘n！中只要把每个数都拆分为质因数的积，则只需要//质因数5的个数即可，因为2的个数会比5多很多，所以质因数有多少5，//尾部就有多少0........#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint num(int n){int cnt=0;while(n){cnt+=n/5;//n/=5;}return cnt;}void find(int n){int left=0,right=INF,mid;while(left<=right){mid=(left+right)/2;if(num(mid)>=n)right=mid-1;elseleft=mid+1;}if(num(left)==n)   printf("%d\n",left);elseprintf("No solution\n");}int main(){int n;while(scanf("%d",&n)!=EOF){if(n==0)printf("1\n");elsefind(n);}return 0;}

 

 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int fun(int x){int sum=0;while(x){sum+=x/5;x/=5;}return sum;}int main(){int m;while(scanf("%d",&m)!=EOF){if(m==0){printf("1\n");continue;}int n=m*4/5*5;while(fun(n)<m){n+=5;}if(fun(n)==0)printf("%d\n",n);elseprintf("No solution\n");}return 0;}