poj 3421 --X-factor Chains(数学、组合)

 

http://poj.org/problem?id=3421

 

X-factor Chains

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4495 Accepted: 1379

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

23410100

Sample Output

1 11 12 12 24 6

Source

POJ Monthly--2007.10.06, ailyanlu@zsu

 

 

 

解法：

s=p1^a1*p2^a2*p3^a3.....pn^an, 其中p1..pn为素数

因为素数不可再分，所以最大值为  (a1+a2+...+an)的最大值。

所以就看s能分解成几个素数的k次方。

所求得的数利用排列组合算下组合数即为最大长度的数量

组合数为：(a1+a2...an)!/(a1!*a2!...*an!)

 

 

 

/**************************************************核心原理：*s=p1^a1*p2^a2*p3^a3.....pn^an, 其中p1..pn为素数**************************************************/#include<cstdio>#include<cstring>#include <cmath>const int M=(1<<20)+5;bool is[M];int prim[M>>1];int a[100],k;int getprm(int n){int i, j, k = 0;int s, e = (int)(sqrt(0.0 + n) + 1);memset(is, 1, sizeof(is));prim[k++] = 2; is[0] = is[1] = 0;for (i = 4; i < n; i += 2) is[i] = 0;for (i = 3; i < e; i += 2) if (is[i]) {prim[k++] = i;for (s = i * 2, j = i * i; j < n; j += s)is[j] = 0;// 因为j是奇数，所以+奇数i后是偶数，不必处理！}for ( ; i < n; i += 2) if (is[i]) prim[k++] = i;return k; // 返回素数的个数}int main(){int i,t,k,sum1,j,n;__int64 sum2;t=getprm(M);while(scanf("%d",&n)==1){if(is[n]){printf("1 1\n");continue;}k=0;for(i=0;i<t;i++){if (n==1) break;if (n%prim[i]==0){a[k]=0;while(n%prim[i]==0) n/=prim[i],a[k]++;//p[k]=count(ak)k++;}}sum1=0;for(i=0;i<k;i++)//sum1 = a1+a2..+ansum1+=a[i];sum2=1;for(i=1;i<=sum1;i++) sum2*=i;for(i=0;i<k;i++)//sum2 = (a1+a2...an)!/(a1!*a2!...*an!) for(j=2;j<=a[i];j++)sum2/=j;printf("%d %I64d\n",sum1,sum2);}return 0;}