如何将整数表示为其质因数的乘积？

问题描述

//Determine the prime factors of a number
for(i = 2; i <= num; i++) {             //Loop to check the factors.
    while(num % i == 0) {               //While the input is divisible to "i" which is initially 2.
        printf("%d ",i);               //Print the factor.
        num = num / i;                  //Divide the num by "i" which is initially 2 to change the value of num.
        }
    }

我知道这是使用 for 循环查找数的质因数的方法。但我不知道如何将输出整数表示为其质因数的乘积。 例如，输入是：10 || 输出是：2 x 5 = 10。我们如何做到这一点？ TIA。

解决方法

你应该：

• 保存原始值。
• 在每个质因数之间打印运算符 x。
• 在最后打印原始值。

#include <stdio.h>

int main(void) {
    int num;
    int i;
    int start_num;
    int is_first = 1;
    if(scanf("%d",&num) != 1) return 1;
    start_num = num;                        //Save the original value.

    //Determine the prime factors of a number
    for(i = 2; i <= num; i++) {             //Loop to check the factors.
        while(num % i == 0) {               //While the input is divisible to "i" which is initially 2.
            if(!is_first) printf("x ");     //Print the operator before second and later operands.
            printf("%d ",i);               //Print the factor.
            num = num / i;                  //Divide the num by "i" which is initially 2 to change the value of num.
            is_first = 0;                   //Mark that there is already one or more operand.
        }
    }
    printf("= %d\n",start_num);            //Print the original value.
    return 0;
}

,

我已经修改了代码，以提供比我之前发布的更健壮一些的东西，同时也稍微提高了效率。我再次假设您希望通过以下范围内的标准输入进行（无符号）32 位输入：SQLQuery sqlQuery = hibernateTemplate.getSession().createSQLQuery(query); sqlQuery.setParamer(...) // set the parameters you need List result = sqlQuery.list();

就算法而言，显然搜索因子只需要测试最多 [1,2^32 - 1] 的候选。还有具有多重性的因子，例如，floor(sqrt(num))。

此外，在分解出 (2) 之后，只需要测试奇数 个因素。

对于 32 位（无符号）类型，质因数将少于 (32) 个。这为用于存储连续素数因子的固定大小数组提供了一个简单的上限。根据使用的算法，数组中的质因子按升序排列。

---

(24) => {2,2,3}

---

,

您可以使用适当的标点符号输出因子：

// Output the prime factors of a number
void factorize(int num) {
    int n = num;                         // save the initial value of num
    const char *sep = "";                // initial separator is an empty string
    for (int i = 2; i <= num / i; i++) { // stop when num is reduced to a prime
        while (num % i == 0) {           // while the input is divisible to "i"
            num = num / i;               // divide the num by "i" (remove the factor)
            printf("%s%d",sep,i);      // print the separator and the factor.
            sep = " x ";                 // change the separator for any further factors
        }
    }
    if (num > 1 || n <= 1) {
        printf("%s%d",num);          // print the last or single factor.
    }
    printf(" = %d\n",n);                // print the rest of the equation
}