4.12训练_P4

Problem Description

Definition 1. n!!…! = n(n−k)(n−2k)…(n mod k), if k doesn’t divide n; n!!…! = n(n−k)(n−2k)…k, if k divides n (There are k marks ! in the both cases).

Definition 2. X mod Y a remainder after division of X by Y.

For example, 10 mod 3 = 1; 3! = 321; 10!!! = 10741.

Given numbers n and k we have calculated a value of the expression in the first definition. Can you do it as well?

Input

contains the only line: one integer n, 1 ≤ n ≤ 10, then exactly one space, then k exclamation marks, 1 ≤ k ≤ 20.

Output

contains one number n!!…! (there are k marks ! here).

郁闷，发了半天没成功，解释就不敲了，大家估计都懂。

#include<cstring>
#include<cstdio>
#include<iostream>
using namespace std;

int main()
{
    char s[30];
    int n,k,t,ans;
    while(scanf("%d %s",&n,s)!=EOF)
    {
        k=strlen(s);
        if(n%k==0)
        {
            for(ans=1;n!=k;n-=k)
                ans*=n;
            ans*=n;
        }
        else
        {
            for(ans=1,t=n%k;n!=t;n-=k)
                ans*=n;
            ans*=t;
        }
        printf("%d\n",ans);
    }
    return 0;
}