poj 2992 Divisors 筛法 算术基本定理

poj 2992 Divisors 筛法 算术基本定理Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11340		Accepted: 3376

DescriptionYour 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?InputThe 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.OutputFor 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 263 - 1.Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

题意:求一个组合数的因子个数

思路:算术基本定理;

#include<iostream>#include<cmath>#include<cstring>#include<queue>#include<cstdio>#define LL long long#define M 432using namespace std;struct m{int a[83];} q[432];bool visit[M];int p[M];int np=0;void ddd(){memset(visit,1,sizeof(visit));for(int i=2; i<M; ++i){if(visit[i]==1){p[np++]=i;for(int j=i*i; j<M; j+=i){visit[j]=0;}}}}void iii(){int t,i,w,j,d;for(i=0; i<83; i++)q[0].a[i]=0;for(i=1; i<=432; i++){t=i;for(w=0; w<=82; w++)q[i].a[w]=q[i-1].a[w];for(j=0; p[j]<=i; j++){while(t%p[j]==0){q[i].a[j]++;t/=p[j];}if(t==1) break;}}}int main(){ddd();iii();LL  ans;int i,b,t,a;while(~scanf("%d %d",&a,&b)){t=a-b;ans=1;int x=a>82?82:a;for(i=0; i<=x; i++)ans*=(1+(long long )q[a].a[i]-q[b].a[i]-q[t].a[i]);printf("%lld\n",ans);}return 0;}