poj 3421 X-factor Chains 组合数学

给定一个数X.

1=X0, X1, X2.....Xm = X 是X的因数

求一串因数,要求Xi | Xi+1,即上一个因数能整除下一个因数,

问这条串就的最长长度,和有多少条这样长度的串.

X = p1^a1 * p2^a2 ... pn^an

Xi = p1^b2 * p2^b2 ...pk^bk... pn^bn,

Xi+1 = p1^b2 * p2^b2 ...pk^(bk+1)... pn^bn,

 

要使length最长,只要从1开始，每次只乘以X的一个质因数即可,即length = (a1+a2+...an)

而方法数就是X的质因数的重排列数,way = (a1+a2+...an)!/(a1!a2!...an!)

/*
 * poj3421.c
 *
 *  Created on: 2011-10-10
 *      Author: bjfuwangzhu
 */

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL long long
#define nmax 1500
int prime[nmax], num[nmax];
int plen, nlen;
void mkprime() {
    int i, j;
    memset(prime, -1, sizeof(prime));
    for (i = 2; i < nmax; i++) {
        if (prime[i]) {
            for (j = i + i; j < nmax; j += i) {
                prime[j] = 0;
            }
        }
    }
    for (i = 2, plen = 0; i < nmax; i++) {
        if (prime[i]) {
            prime[plen++] = i;
        }
    }
}
void solve(int n) {
    int i, j, te, cnt, res1;
    LL res2;
    te = (int) sqrt(n * 1.0);
    for (i = 0, nlen = 0; (i < plen) && (prime[i] <= te); i++) {
        if (n % prime[i] == 0) {
            cnt = 0;
            while (n % prime[i] == 0) {
                n /= prime[i];
                cnt++;
            }
            num[nlen++] = cnt;
        }
    }
    if (n > 1) {
        num[nlen++] = 1;
    }
    for (i = 0, res1 = 0; i < nlen; i++) {
        res1 += num[i];
    }
    for (i = 2, res2 = 1; i <= res1; i++) {
        res2 = res2 * i;
    }
    for (i = 0; i < nlen; i++) {
        for (j = 2; j <= num[i]; j++) {
            res2 = res2 / j;
        }
    }
    printf("%d %lld\n", res1, res2);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int n;
    mkprime();
    while (~scanf("%d", &n)) {
        solve(n);
    }
    return 0;
}