计算有序和无序排列 - C答案

【问题标题】：calculating ordered and unordered permutation - C计算有序和无序排列 - C
【发布时间】：2015-11-03 01:57:33
【问题描述】：

尝试计算 C 中数字和选择的有序排列和无序排列。递归阶乘函数正在工作，排列结果给出随机内存位置

./排列
输入项目数：5
输入要选择的数字：2
有1.#INF00有序排列
有1.#INF00个无序排列

#include <stdio.h>

double ordered_permutation(double total_items, double selection);
double unordered_permutation(double total_items, double selection);
double factorial( double number);

int main() {

    double items;
    double selection;
    printf("Enter number of items: ");
    scanf("%f", &items);
    printf("Enter number to select: ");
    scanf("%f", &selection);

    double ordered_perm;
    double unordered_perm;
    ordered_perm = ordered_permutation( items, selection );
    unordered_perm = unordered_permutation( items , selection );

    printf("There are %f ordered permutations\nThere are %f unordered permutations\n", ordered_perm, unordered_perm);

    return 0;
}

/*    total_items!/ ( total_items! - selection! )    */
double ordered_permutation(double total_items, double selection){
    double permutations;

    permutations = factorial(total_items)/ ( factorial(total_items) - factorial(selection) );

    return permutations;
}

/*    total_items!/ ( ( total_items! - selection! ) selection! )    */
double unordered_permutation(double total_items, double selection){
    double permutations;

    permutations = factorial(total_items)/ ( ( factorial(total_items) - factorial(selection) ) * factorial(selection));

    return permutations;
}

double factorial( double number){
    if( number <= 1 )
        return 1;
    else 
        return  number * factorial((number-1));
}

【问题讨论】：

标签： c recursion combinations permutation combinatorics

【解决方案1】：

将分母调用更改为：

    factorial(total_items - selection)

和

    (factorial(total_items- selection) * factorial(selection))

【讨论】：

【解决方案2】：

这不是您问题的完全答案，但使用递归计算大阶乘可能会导致堆栈溢出 [并且相对] 缓慢。

此外，您将double 用于计数和索引。阶乘是基于整数相乘，结果只能是整数。

这是一个非递归阶乘函数：

#if 1
typedef int value_t;
#else
typedef long long value_t;
#endif

value_t
factorial(value_t n)
{
    value_t i;
    value_t f;

    f = 1;

    for (i = 1;  i <= n;  ++i)
        f *= i;

    return f;
}

【讨论】：