[C] acquaintance of a Exercises

Given the value 1. The two integer variables, the contents of the two values to be exchanged.
This topic and ideas: the definition of temporary variables c

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int a = 3, b = 4, c;  //定义变量
    printf("初始值：%d %d\n", a, b);
    c = a; 
    a = b; 
    b = c;
    printf("交换后：%d %d\n", a, b);
    system("pause");
    return 0;
}

Execution results are as follows:

2 does not allow the creation of temporary variables, swap the contents of two numbers (additional topic)
This topic ideas: the use of the difference between a and b are the exchange operation

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int a = 3, b = 4;
    printf("初始值：%d %d\n", a, b);
    a = a + b;
    b = a - b;
    a = a - b;
    printf("交换后：%d %d\n", a, b);
    system("pause");
    return 0;
}

Execution results are as follows:

3. The maximum value of 10 integers requirements.
This question ideas: find the maximum without defined pointer variable.
Note: It is recommended not to use arrays compare apples to apples, because the array index must be a positive integer or a logical value.

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main()
{
    int num, max=0x80000000;//给max附一个最小值
    for (int i = 0; i < 10; i++){
        scanf("%d", &num);
        if (num>max)
            max = num;
    }
    printf(" %d\n", max );
    system("pause");
    return 0;
}

Performed as follows:

4. The three numbers in descending output
this question ideas : pairwise comparison, to ensure a large exchange in the small front

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main()
{
    int a, b, c, m;
    scanf(" %d %d %d" ,&a,&b,&c);
    if (a<b)
    {
        m = a;  a = b; b = a;
    }
    if (a<c)
    {
        m = a; a = c; c = a;
    }
    if (b<c)
    {
        m = b; b = c; c = b;
    }
    printf("从大到小排序为:%d %d %d\n", a, b, c);
    system("pause");
    return  0;
}

Performed as follows:

The greatest common divisor of two numbers.
Outline of Solution 1: for i traversed, whether simultaneously be divisible by two integers (slower)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main()
{  
    int a=0, b=0, max=0;
    scanf("%d %d", &a, &b);
    int smaller;
    smaller = a > b ? b : a;
    for (int i = 2; i <= smaller; i++){
        if (a%i == 0 && b%i == 0){
            max = i;
        }   
    }
    printf("最大公约数：%d\n", max);
    system("pause");
    return 0;
}