Divisors pku 2992

Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6789		Accepted: 1783

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 2 ⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

Source

CTU Open 2005

#include<stdio.h> int p[83]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431}; int d[432][83]={0},m=0; void divisors(int n) { int i=n,j; for (j=0;n>1;j++) { while (n%p[j]==0) { d[i][j]++; n/=p[j]; } } if (j>m) m=j; } int main() { int i,k,l,n; __int64 j; for (i=2;i<432;i++) { for (j=0,k=i-1;j<m;j++) d[i][j]=d[k][j]; divisors(i); } while (scanf("%d%d",&n,&k)!=EOF) { l=n-k; for (i=0,j=1;i<83;i++) j*=(d[n][i]-d[k][i]-d[l][i]+1); printf("%I64d/n",j); } return 0; }