c - 质因数之和

Question

我正在尝试显示给定数字的质因数之和我在输出中显示主要因素时遇到困难。

示例输出:

Input number: 6
Factors are: 1 2 3
Sum of its factor: 1 +2 +3 =6

我能够显示总和，但我想显示 1+2+3=6，就像上面示例中的因数是 1 2 3 一样。

你能帮我纠正我的语法来实现这个吗？提前致谢。

这是我的代码:

#include <stdio.h>

int main() {
    int i, j, num, isPrime, sum;

    printf("Input number: ");
    scanf("%d", &num);

    printf("Factors are: ", num);

    for (i = 1; i <= num; i++) {
        if (num % i == 0) {
            isPrime = 1;
            for (j = 2; j <= i / 2; j++) {
                if (i % j == 0) {
                    isPrime = 0;
                    break;
                }
            }

            if (isPrime == 1) {
                printf("%d ", i);
                sum += i;
            }
        }
    }

    printf("\nSum of its factor : %d", sum);
    return 0;
}

Answer 1

您的代码实际上具有未定义的行为，因为 sum 未初始化为 0。它只是偶然产生正确的总和。

您可以将因子存储在数组中，甚至可以在使用 sprintf 时构建表达式。表达式的最大长度不是很大，因为最多可以有 9 个不同的素因子 (29!! > 2³²)

修改后的版本:

#include <stdio.h>

int main() {
    char expr[9 * 11 + 1];
    int i, j, pos, num, isPrime, sum;

    printf("Input number: ");
    if (scanf("%d", &num) != 1)
        return 1;

    printf("Factors are: 1");  // always include 1
    pos = 0;
    expr[pos] = '\0';
    sum = 1;
    for (i = 2; i <= num; i++) {
        if (num % i == 0) {
            isPrime = 1;
            for (j = 2; j * j <= i; j++) {
                if (i % j == 0) {
                    isPrime = 0;
                    break;
                }
            }
            if (isPrime == 1) {
                pos += sprintf(expr + pos, "+%d", i);
                printf(" %d", i);
                sum += i;
            }
        }
    }

    printf("\nSum of its factors: 1%s = %d\n", expr, sum);
    return 0;
}

输出:

Input number: 6
Factors are: 1 2 3
Sum of its factors: 1+2+3 = 6

这是一个更健壮和更快的版本，对于非常大的 num 值没有未定义的行为:

#include <stdio.h>

int main() {
    char expr[9 * 11 + 1];
    int i, pos, num;
    unsigned sum;

    printf("Input number: ");
    if (scanf("%i", &num) != 1)
        return 1;

    printf("Factors are: 1");  // always include 1
    pos = 0;
    expr[pos] = '\0';
    sum = 1;
    for (i = 2; num / i >= i; i++) {
        if (num % i == 0) {
            pos += sprintf(expr + pos, "+%d", i);
            printf(" %d", i);
            sum += i;
            do { num /= i; } while (num % i == 0);
        }
    }
    if (num != 1) {
        pos += sprintf(expr + pos, "+%d", num);
        printf(" %d", num);
        sum += num;
    }
    printf("\nSum of its factors: 1%s = %u\n", expr, sum);
    return 0;
}

测试:

Input number: 0x7fffffff
Factors are: 1 2147483647
Sum of its factors: 1+2147483647 = 2147483648