POJ2992 Divisors 組合數，分解質因數
阿新 • • 發佈：2019-02-09
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 263 - 1.
Sample Input
5 1
6 3
2
6
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 263 - 1.
Sample Input
5 1
6 3
10 4
Sample Output2
6
16
這道題是求組合數的約數個數，利用c(n,k)=(n-k+1)/k*c(n,k-1) 進行素因子的遞推即可，不需要高精度。
#include <iostream> #include <cstdio> #include <map> #include <cmath> #include <map> #include <vector> #include <algorithm> #include <cstring> #include <string> using namespace std; #define LL long long #define maxn 1001 int a[maxn]; LL f[maxn][maxn]; int prime[100],c; int v[maxn]; void p() { int i,j,n=maxn,m; c=0; m=(LL)sqrt(n+0.5); memset(v,0,sizeof(v)); for(i=2;i<=m;i++) if(!v[i]){ for(j=i*i;j<=n;j+=i) v[j]=1; } for(j=2;j<=n;j++){ if(!v[j]){ prime[c++]=j; } } } void ad(int n,int flag) { if(flag==1){ for(int i=0;i<c&&n>1;i++){ while(n%prime[i]==0){ n/=prime[i]; a[i]++; } } }else{ for(int i=0;i<c&&n>1;i++){ while(n%prime[i]==0){ n/=prime[i]; a[i]--; } } } } void pr(int i,int j) { f[i][j]=1; for(int t=0;t<c;t++){ f[i][j]*=(a[t]+1); } } void fen() { for(int i=0;i<=431;i++){ memset(a,0,sizeof(a)); f[i][0]=1; for(int j=1;j<=i;j++){ ad(i-j+1,1); ad(j,0); pr(i,j); } } } int main() { int n,k; p(); fen(); while(~scanf("%d%d",&n,&k)){ printf("%lld\n",f[n][k]); } return 0; }