POJ 3421

X-factor Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5111		Accepted: 1622

Description

Given a positive integer X, an X-factor chain of length m is a sequence of integers,

1 = X₀, X₁, X₂, …, X_m = X

satisfying

X_i < X_i+1 and X_i | X_i+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

 

找出x的所有质因子表达式，最大长度就是各质因子指数的和，然后按照排列组合计算即可。

 

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>

 using namespace std;

 #define maxn (1 << 20) + 5

 typedef long long ll;

 int x,len = ;
 bool prime[maxn];
 int ele[];

 ll cal(int x) {
         ll ans = ;
         for(int i = ; i <= x; i++) ans *= i;

         return ans;
 }

 void init() {
         for(int i = ; i <= maxn - ; i++) {
                 prime[i] = i %  || i == ;
         }

         for(int i = ; i * i <= maxn - ; i++) {
                 if(!prime[i]) continue;
                 for(int j = i; j * i <= maxn - ; j++) {
                         prime[j * i] = ;
                 }
         }

         len = ;
         for(int i = ; i <= maxn - ; i++) {
                 if(prime[i])
                 ele[len++] = i;
         }

 }
 int main() {
         //freopen("sw.in","r",stdin);

         init();

         while(~scanf("%d",&x)) {
                 if(prime[x]) {
                         printf("1 1\n");
                 } else {
                         int ans1 = ,len2 = ;
                         ll ans2 = ;
                         int num[];
                         for(int i = ; i < len && ele[i] <= x && x != ; i++) {
                                 int sum = ;
                                 if(x % ele[i] == ) {
                                         while(x % ele[i] == ) {
                                                 ans1++;
                                                 sum++;
                                                 x /= ele[i];
                                         }
                                         num[len2++] = sum;
                                 }

                         }

                         ans2 = cal(ans1);
                         for(int i = ; i < len2; i++)  {
                                 ans2 /= cal(num[i]);
                         }
                         printf("%d %I64d\n",ans1,ans2);

                 }

         }

         return ;
 }