仅打印总和为10的3位数字的数组 [英] Print only those arrays with 3 digits whose sum is 10 - C program

问题描述

输出：

1 2 3 4 
1 2 7 
1 3 6 
1 4 5 
1 9 
2 3 5 
2 8 
3 7 
4 6 
10 

预期输出：

1 2 7 
1 3 6 
1 4 5 
2 3 5 

I只希望总和为10且也只有3位的那几对数字，即应该显示总和为10的3对不同的数字，其余所有对都跳过或不显示。

I want only those pairs of digits whose sum is 10 and also has only 3 digits i.e, the pair of 3 distinct digits whose sum is 10 should be displayed, rest all other pairs get skip or not displayed.

下面是我为解决此问题而编写的完整源代码。

Below is my full source code which I have written for this problem.

#include <stdio.h>
#include <stdlib.h>

void partition(int part) {
    int *parts, *ptr, i, idx = 0, tot = 0, cur = 1, max = 1;   

    while ((max * (max + 1)) / 2 <= part)
        max++;`

    ptr = parts = malloc(sizeof(int) * max);

    for (;;) {
        if ((tot += *ptr++ = cur++) < part)
            continue;

        if (tot == part) {
            for (i = 0; i < ptr-parts; i++) {
                printf("%d ", parts[i]);
            }
            printf("\n");
        }

        do {
            if (ptr == parts) {
                free(parts);
                return;
            }
            tot -= cur = *--ptr;
        } while (++cur + tot > part);
    }
}

int main(int argc, char*argv[]) {
    int n;
    scanf("%d", &n);
    partition(n);
    return 0;
}

推荐答案

您的代码似乎太复杂了。这是一个简单的解决方案：

Your code seems way too complicated. Here is a simple solution:

#include <stdio.h>

#define SUM 10

int main(void) {
    for(int a = 1; a < SUM; a++) {
        for(int b = a + 1; b < SUM; b++) {
            for(int c = b + 1; c < SUM; c++) {
                if(a + b + c == SUM) {
                    printf("%d %d %d\n", a, b, c);
                }
            }
        }
    }
}

程序输出：

1 2 7
1 3 6
1 4 5
2 3 5

这可能更有效，但是它是一种简单的形式。

This could be more efficient but it is a simple form.

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