X-factor Chains(poj3421）（素数因子）

题解：就是给你一个正整数，然后你求出他的因字串从小到大排列的最大长度，以及这个最大长度的种类数

所以就找所有的素数因子，素数因子数即为最大长度数，然后就是，这几个因字数的全排列，即为种类数（注意有重复因子，这就用到数学知识了，需要去重）

Language:Default X-factor Chains Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9320   Accepted: 2981 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 Source POJ Monthly--2007.10.06, ailyanlu@zsu #include<stdio.h> #include<map> #include<algorithm> using namespace std; typedef long long ll; ll mult(int a) { ll d=1; for(int i=2;i<=a;i++) { d=d*i; } return d; } int main() { int n; while(scanf("%d",&n)!=EOF) { map<int,int>ma; int t=n; int i; for(i=2;i*i<=n;i++) { while(t%i==0&&t) { t=t/i; ma[i]++; } } if(t>1) ma[t]++; map<int,int>::iterator iter; int len=0; for(iter=ma.begin();iter!=ma.end();iter++) { len+=iter->second; } ll ans=mult(len); for(iter=ma.begin();iter!=ma.end();iter++) { ans=ans/mult(iter->second); } printf("%d %lld\n",len,ans); } return 0; }