Divisors
Time Limit:
1000MS
Memory Limit: 65536K
Description
Your task in this problem is to determine the number of divisors of
Cnk. 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
Cnk. 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
题目大意:求c(n,m)的因子个数
思路:假设将一个数表示成它的质因数分解,如A=a^p1*b^p2*c^p3*...*n^pn.
那么它的约数个数就是:ans=(p1+1)*(p2+1)*(p3+1)*...*(pn+1).
而C(n,k)=n!/[(k!*(n-k)!],c[n][k]代表n的阶乘时能够分解出几个k。
那么只需要求出他们的阶乘对于每一个素数的个数就可以了。
公式:ai=c[n][prime[i]]-c[k][prime[i]]-c[(n-k)][prime[i]]。ans=a1*a2.*...*ak (k代表当prime[k]小于n的时候)。
#include <stdio.h>
using namespace std;
const int maxn = 432;
bool vis[maxn];
int num[maxn][maxn];
int main()
{
int n,k,i,j,t;
//判断素数
for(i=2;i<22;i++)
{
if(!vis[i])
{
for(j=i*i;j<maxn;j+=i)
vis[j]=true;
}
}
//质因数计数公式
for(i=2;i<maxn;i++)
{
for(j=0;j<=i;j++)
num[i][j]=num[i-1][j];
t=i;
for(j=2;j<maxn&&t>1;j++)
{
if(!vis[j])
{
while(t%j==0)
{
num[i][j]++;
t/=j;
}
}
}
}
while(~scanf("%d%d",&n,&k))
{
__int64 ans=1;
if(k&&n-k)//k==0||(n-k)==0时,Cnk为1
{
for(i=2;i<=n;i++)
{
if(!vis[i])
ans*=num[n][i]-num[k][i]-num[n-k][i]+1;
}
}
printf("%I64d\n",ans);
}
return 0;
}