[C language] Enter the year, month, and day to find the day of the year? And judge the leap year, if it is a leap year, return 1, if not, return 0

code show as below

//输入年月日求是该年的第几天？并判断闰年，若是闰年返回1，不是则返回0
#include <stdio.h>
int LeapYear(int year);//判断是不是闰年的函数。函数原型声明 
int main()
{
    
    
	int year,month,day;
	printf("请输入日期:\n");
	
	while(1) //控制用户输入格式，输入的 月份和天数 正确才可继续，否则需要重新输入 
	{
    
    
		scanf("%d,%d,%d",&year,&month,&day);
		if(month<1 || month>12 || day<1 || day>31)
			printf("输入日期错误！请您重新输入！\n");	
		else 
			break;	//若正确，则跳出while循环，继续执行下面步骤 
	}
	
	//下面数组中的0值得思考。
	//可以这么想，如果用户输入2020,1,15。（假设用户输入正确） 
	//那么由于是1月份，所以用户输入的第三个数字 是 几 就是第几天 
	int array[]={
    
    0,28,31,30,31,30,31,31,30,31,30,31};
	
	int sum=0;  //sum即为该年的"第多少天" 
	if(month==1)
		sum=day;
	else if(month==2)//若是2月份,"第多少天"应该再加上前边1月份的天数 
		sum=day+31;
	else{
    
    //若既不是1月份，也不是2月份，那么开始求是"第多少天" 
		for(int i=0;i<month;i++)
			sum+=array[i];//比如：2018,3,23。那么即:array[0]+array[1]+array[2]=0+28+31。还应再 +23。最后=第82天 
		sum+=day;//因为下标是从0开始的，故应再添加上输入的天数(日子) 
		
		if(LeapYear(year)) //若是闰年，算出来的天数应该再加上1 
			sum++;
	}
	
	printf("该日期是第%d天\n",sum);
	return 0;
}

int LeapYear(int year)
{
    
    
	if((year%4==0 && year%100!=0) || (year%400==0))
	{
    
    
		return 1;
	}
	else 
		return 0;
}

test

The code analysis is actually written very clearly in the comments!

Below are a few test inputs and their output results.

This example is what I wrote in the comments:

2018 is not a leap year, so +1 is not needed.

Here's another example of testing leap years:

Obviously, compared to the previous result, in this test, since 2020 is a leap year, when a day in March is the day of the year, February has 29 days, so the final result should be 82 +1=83.

Next, test the prompt of the user error output:

Here's what happens when the months entered don't match:

Try again with a different month and year:

Here's what happens when the number of days is wrong:

Because only the changed output statement, no longer put all the code, there are ah!

printf("输入日期错误！请您重新输入:\n");
printf("该日期是%d年的第%d天\n",year,sum);

A few tests:

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There is a small bug, I wonder if you found it:

...So, it is also a "pot" for input judgment, haha! I'll leave it to the reader to improve it! ! !