POJ 2992 Divisors（求组合数因子个数）

Divisors

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12048 Accepted: 3594

Description

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 2⁶³ - 1.

Sample Input

5 16 310 4

Sample Output

2616

Source

CTU Open 2005

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题意：求C(n,k)的因子个数。

题解：C(n,k)=n!/k!/(n-k)!，直接将所有阶乘的数分解出所有素因数然后统计一下

然后套用因数个数公式就好   (p1+1)*(p2+1)*(p3+1)*...*(pn+1)

题目很简单，然而卡时间，500*50*50都卡？？

好吧，预处理一下就OK了。。。。。

定义dp[i][j]：表示i的阶乘里有几个素数j。

#include<set>      #include<map>         #include<stack>                #include<queue>                #include<vector>        #include<string>     #include<time.h>    #include<math.h>                #include<stdio.h>                #include<iostream>                #include<string.h>                #include<stdlib.h>        #include<algorithm>       #include<functional>        using namespace std;                #define ll long long          #define inf 1000000000           #define mod 1000000007                #define maxn  50500    #define lowbit(x) (x&-x)                #define eps 1e-9  ll a[505]={1,1},b[500];ll dp[450][450],cnt;void init(){ll i,j;for(i=2;i<=432;i++){if(a[i])continue;b[++cnt]=i;for(j=i*i;j<=432;j+=i)a[j]=1;}for(i=2;i<=432;i++){for(j=1;b[j]<=i && j<=cnt;j++){ll tmp=i,res=0;while(tmp)tmp/=b[j],res+=tmp;dp[i][b[j]]=res;}}}int main(void){init();ll i,x,y;while(scanf("%lld%lld",&x,&y)!=EOF){ll ans=1;for(i=1;b[i]<=x && i<=cnt;i++){ll tmp=dp[x][b[i]]-dp[y][b[i]]-dp[x-y][b[i]];ans=ans*(tmp+1);}printf("%lld\n",ans);}return 0;}