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UVA_10139

2018-06-17 21:35:19来源：未知阅读 ()

The factorial function, n! is de?ned thus for n a non-negative integer:
0! = 1 n! = n×(n?1)! (n > 0)
We say that a divides b if there exists an integer k such that k×a = b

Input
The input to your program consists of several lines, each containing two non-negative integers, n and m, both less than 231.

Output
For each input line, output a line stating whether or not m divides n!, in the format shown below.

Sample Input
6 9

6 27

20 10000

20 100000

1000 1009

Sample Output
9 divides 6!

27 does not divide 6!

10000 divides 20!

100000 does not divide 20!

1009 does not divide 1000!

m能否被n!整除，题目并不难，细节扣的多。分解m的质因子，然后通过勒让德的结论就能过。

while（n）{

　　n/=x;

　　sum+=n;

}

被英语语法gank了一波，还要注意的是 0和1 也能整除。

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define N 1000010
using namespace std;
int vis[N];
int prime[N];
int pn=0,flag;
void gp()
{
    for (int i = 2; i <= N; i++) {
        if (vis[i]) continue;
        prime[pn++] = i;
        for (int j = i; j <= N; j += i)
            vis[j] = 1;
    }
}
int lrd(int n,int x)
{
    int sum=0;
    while(n)
    {
        n/=x;
        sum+=n;
    }
    return sum;
}
int main()
{
    gp();
    //cout<<prime[pn-1]<<endl;
    int m,n;
    while(~scanf("%d%d",&n,&m))
    {
        if((n>=m)||(n==0&&m==1))
        {
             printf("%d divides %d!\n",m,n);
             continue;
        }
        flag=0;
        int x=m;
        for(int i=0;prime[i]*prime[i]<=m;i++)
        {
            int cnt=0;
           while(x%prime[i]==0)
           {
               x/=prime[i];
               cnt++;
           }
           if(cnt)
           {
               int res=lrd(n,prime[i]);
               if(res<cnt)
                    flag=1;
           }
        }
        if((x<=n&&!flag))
            printf("%d divides %d!\n",m,n);
        else
            printf("%d does not divide %d!\n",m,n);
    }
}

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