C language daily practice (2)

C Practice Example 4

Topic: Enter a certain year, month, and day, and judge whether this day is the day of the year?

Program Analysis: Taking March 5th as an example, you should first add up the previous two months, and then add 5 days, which is the day of the year. In special cases, when the input month is greater than 3 in a leap year, you need to consider adding an extra day .

Program source code:

#include <stdio.h>
int main()
{
    int day,month,year,sum,leap;
    printf("\n请输入年、月、日，格式为：年,月,日（2015,12,10）\n");
    scanf("%d,%d,%d",&year,&month,&day);  // 格式为：2015,12,10
    switch(month) // 先计算某月以前月份的总天数
    {
        case 1:sum=0;break;
        case 2:sum=31;break;
        case 3:sum=59;break;
        case 4:sum=90;break;
        case 5:sum=120;break;
        case 6:sum=151;break;
        case 7:sum=181;break;
        case 8:sum=212;break;
        case 9:sum=243;break;
        case 10:sum=273;break;
        case 11:sum=304;break;
        case 12:sum=334;break;
        default:printf("data error");break;
    }
    sum=sum+day; // 再加上某天的天数
    if(year%400==0||(year%4==0&&year%100!=0)) {// 判断是不是闰年
        leap=1;
    } else {
       leap=0;
    }
    if(leap==1&&month>2) { // *如果是闰年且月份大于2,总天数应该加一天
        sum++;
    }
    printf("这是这一年的第 %d 天。",sum);
    printf("\n");
}

The output of the above example is:

Please enter the year, month, and day in the format: year, month, day (2015,12,10) 
2015,10,1 
which is the 274th day of the year.

C Practice Example 5

Topic: Input three integers x, y, z, please output these three numbers from small to large.

Program analysis: We find a way to put the smallest number on x, first compare x with y, if x>y, exchange the values ​​of x and y, and then compare x with z, if x> then Exchange the values ​​of zx and z so that x is minimized.

Program source code:

#include <stdio.h>

int main()
{
    int x,y,z,t;
    printf("\n请输入三个数字:\n");
    scanf("%d%d%d",&x,&y,&z);
    if (x>y) { /*交换x,y的值*/
        t=x;x=y;y=t;
    }
    if(x>z) { /*交换x,z的值*/
        t=z;z=x;x=t;
    }
    if(y>z) { /*交换z,y的值*/
        t=y;y=z;z=t;
    }
    printf("从小到大排序: %d %d %d\n",x,y,z);
}

The output of the above example is:

Please enter three numbers: 
1 
3 
2 
Sort from small to large: 1 2 3