X-factor Chains
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7986 Accepted: 2546
Description
Given a positive integer X, an X-factor chain of length m is a sequence of integers,
1 = X0, X1, X2, …, Xm = X
satisfying
Xi < Xi+1 and Xi | Xi+1 where a | b means a perfectly divides into b.
Now we are interested in the maximum length of X-factor chains and the number of chains of such length.
Input
The input consists of several test cases. Each contains a positive integer X (X ≤ 220).
Output
For each test case, output the maximum length and the number of such X-factors chains.
Sample Input
2
3
4
10
100
Sample Output
1 1
1 1
2 1
2 2
4 6
将一个数分解成若干个素数相乘,因为后面分解出来的素数一定是在前面分解之后分解的,所以自然就大2倍以上。然后问有多少条链,其实就是问将所有因子的全排列有多少种,但是要除去重复因子的全排列。
n个不同的数的全排列就是n!,除去重复的就是n!除以每个重复因子数的阶乘。
注:防止爆掉都用的long long 存储
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
using namespace std;
typedef long long ll;
const int N=22;
ll fac[N+1]={1};
void init(){
for(ll i=1;i<=N;i++){
fac[i]=fac[i-1]*i;
}
}
int main(){
init();
ll n;
while(cin>>n){
ll sum=0,d=1;
for(ll i=2;i*i<=n;i++){
if(n%i==0){
ll cnt=0;
while(n%i==0){
n/=i;
cnt++;
}
sum+=cnt;
d*=fac[cnt];
}
}
if(n>1) sum++;
cout<<sum<<" "<<fac[sum]/d<<endl;
}
}