TP iostream 約數個數 PE printf for tinc 階乘 BE

Your task in this problem is to determine the number of divisors of C_n^k. Just for fun -- or do you need any special reason for such a useful computation?

Input

The input consists of several instances. Each instance consists of a single line containing two integers n and k (0 ≤ k ≤ n ≤ 431), separated by a single space.

Output

For each instance, output a line containing exactly one integer -- the number of distinct divisors of C_n^k. For the input instances, this number does not exceed 2 ⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16
思路:計算約數個數的時候,可以類比數的唯一分解求約數個數;
s=2^p1*3^p2*5^p3*-----
num=(p1+1)*(p2+1)*(p3+1)*....
如果求分數類型的約數個數...
可以分解質因數之後在上面的個數－下面的個數
即s__=2^(p1-p1_)*3^(p2-p2_)*----
num=(p1-p1_+1)*(p2-p2_+1)*(p3-p3_+1)*---
階乘的質因數分解:
　　int cal(int n,int p) if(n<p)return 0;else return n/p+cal(n/p,p);

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int prime[500];bool vis[500];
typedef long long ll;int cnt;
void init(){
	for(int i=2;i<=431;i++){
		if(!vis[i]) for(int j=i*2;j<=431;j+=i) vis[j]=1;
	}
	for(int i=2;i<=431;i++){
		if(!vis[i]) prime[++cnt]=i;
	}
}
int num1[500],num2[500];
int cal(int n,int p){
	if(n<p) return 0;
	else return n/p+cal(n/p,p);
}
int main(){
	init();int n,m;
	//std::ios::sync_with_stdio(false);cin.tie(0);
	//for(int i=1;i<=83;i++) cout<<prime[i]<<endl;
	while(~scanf("%d%d",&n,&m)){
		ll ans=1;
		memset(num1,0,sizeof(num1));
		memset(num2,0,sizeof(num2));
		for(int i=1;prime[i]<=n&&i<=cnt;i++){
			num1[i]=cal(n,prime[i]);
			num2[i]=cal(m,prime[i])+cal(n-m,prime[i]);
			ans*=(num1[i]-num2[i]+1);
			//cout<<ans<<endl;		
		}
		printf("%lld\n",ans);
	}
	return 0;
} 

　　

poj2992 divisors 求組合數的約數個數,階乘的質因數分解