| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9856 | Accepted: 2896 |
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 computationInput
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
Source
CTU Open 2005求C(n,k)的约数的个数。 n和k都很大，直接求肯定不行。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)。
//3624K 610MS #include#include #define N 100007 bool visit[1010]; long long c[1500][1500]; long long prime[107]; void init_prim()//prime存的是下标，visit存的是数。visit[5]==true。 { long long num=0; memset(visit,true,sizeof(visit)); for(long long i=2;i<1007;++i) { if(visit[i]==true) { num++; prime[num]=i; } for(long long j=1;((j<=num)&&(i*prime[j]<=10007));++j) { visit[i*prime[j]]=false; if(i%prime[j]==0) break; } } } void init() { for(long long i=2;i<=437;i++) for(long long j=1;prime[j]<=i;j++) { long long n=i,res=0; while(n){n/=prime[j];res+=n;} c[i][prime[j]]=res; } } int main() { long long n,k; init_prim(); init(); while(scanf("%I64d%I64d",&n,&k)!=EOF) { long long ans=1,a; for(long long i=1;prime[i]<=n;i++) { a=c[n][prime[i]]-c[k][prime[i]]-c[n-k][prime[i]]; ans*=(1+a); } printf("%I64d\n",ans); } return 0; }