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

Sample Output

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 ;

 }

巴特西

POJ 3421

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