题目链接:
http://poj.org/problem?id=2992
解题思路:
求C(n,k)的因子个数。。。
将1!~431!的每个素因子的个数打表,而C(n,k)=n!/(k!*(n-k)!),所以再利用求因子个数的公式即可http://blog.youkuaiyun.com/piaocoder/article/details/47954385
AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
typedef long long ll;
vector<int> v;
int prime[500];
int a[450][450];
int vis[500],nprime;
int sum[500];
void getprime(){
nprime = 0;
memset(vis,0,sizeof(vis));
for(int i = 2; i <= 450; i++){
int t = 450/i;
for(int j = 2; j <=t; j++)
vis[i*j] = 1;
}
for(int i = 2; i <= 450; i++){
if(!vis[i])
prime[nprime++] = i;
}
}
void div(int x){
v.clear();
int t = x;
for(int i = 0; i < nprime && prime[i]*prime[i] <= x; i++){
while(t % prime[i] == 0){
v.push_back(prime[i]);
t /= prime[i];
}
if(t == 1)
break;
}
if(t > 1)
v.push_back(t);
}
void solve(){
memset(a,0,sizeof(a));
for(int i = 2; i <= 431; i++){
memcpy(a[i],a[i-1],sizeof(a[i]));
div(i);
for(int j = 0; j < v.size(); j++)
a[i][v[j]]++;
}
}
int main(){
getprime();
solve();
int n,k;
while(~scanf("%d%d",&n,&k)){
memset(sum,0,sizeof(sum));
for(int i = 1; i <= 431; i++)
sum[i] += a[n][i]-a[k][i]-a[n-k][i];
ll ans = 1;
for(int i = 2; i <= 431; i++)
ans *= (sum[i]+1);
printf("%lld\n",ans);
}
return 0;
}