对指定日期计算出N天后的日期问题

								    指定日期计算出N天后的日期问题

问题描述：先输入起始日期，再输入任意天数n，求n天后的日期。

源代码一（参考http://zhidao.baidu.com/linkurl=0BxF0QvTywzuRkLUFj6ulUQKYqp97CjUOZBV1UUmXoCkS9HNyOT2axHkcOsRLi-xlB_JE6KvdglSD23S35EUXK）

#include<stdio.h>
<span style="font-family:SimHei;font-size:14px;"><strong>int main()
{
   int year,month,day;
   int leapyear=0;
   int daytime=0;
   int sum;
   static int month_date[12]={31,28,31,30,31,30,31,31,30,31,30,31};
   printf("请输入日期:");
   scanf("%d-%d-%d",&year,&month,&day);
   printf("请输入天数:");
   scanf("%d",&daytime);
   sum=daytime+day;
   do
   {
	   	if(month==2)
	   	{
	   		month_date[month-1]+=leapyear;

	   	}
	   	if(sum>month_date[month-1])
	   	{
	   		sum-=month_date[month-1];
	   		month++;
	   		if(month==13)
	   		{
	   			year++;
	   			if(year%400==0||(year%100!=0&&year%4==0))
	   			{
	   				leapyear=1;
	   			}
	   			else
	   			{
	   				leapyear=0;
	   			}
	   			month=1;
	   		}
	   	}
   } while(sum>month_date[month-1]);
   day=sum;
   printf("\n第%d天后是%d-%d-%d",daytime,year,month,day);
   return 0;
}</strong></span><span style="font-family: arial, 'courier new', courier, 宋体, monospace; font-size: 14px;">
</span>
<span style="font-family:SimHei;font-size:14px;"><strong>
</strong></span>
源代码二（自己写的）

#include<stdio.h>
int year,month,day,days,years;
int a[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};

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

void rest(int days)
{
	int i;
	for(i=1;i<=12;i++)
	{
			if(days-a[i]>0)
				days=days-a[i];
			else
			{
			   	month=i;
				day=days;
			    break;
			}
	}
     if(i-1==12)
	 {
		 year=year+1;
		 if(!ru(year))
			 days++;
		 rest(days);
	 }
	 year+=years;
}

int main()
{
	int i;
	while(scanf("%d%d%d%d",&year,&month,&day,&days)==4)
	{
		days=days-(a[month]-day);
		if(!ru(year)&&month<=2)
			days=days+1;
		for(i=month+1;i<=12;i++)
		{
			if(days>a[i])
				days=days-a[i];
			else
				break;
		}
		if(i-1==12)
		{
			printf("%d\n",days);
			year++;
			years=0;
			while(days>=366)
			{
			   days=days-366;
			   years++;
			}
			for(i=1;i<=years;i++)
			{
				if(!ru(year+i))
					days++;
			}
			rest(days);
		}
		else
		{
				month=i;
				day=days;
		}

		printf("%d-%d-%d\n",year,month,day);
	}
	return 0;
}