【日期】计算后续日期

【日期】计算后续日期

我们经常要计算，从今天往后N天之后是哪一天（哪年哪月哪日）。现在我们就可以编写一个程序，推算指定日期之后的第N天是什么日期。

输入：
年 月 日
N

输出：
（年月日+N天后的）年.月.日

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

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
	int y,m,d,k=0,i,j,l,n,r;
	scanf("%d %d %d",&y,&m,&d);
	int x,p;
	scanf("%d",&x);
	switch(m)
    {  
        case 1:k=d;break;  
        case 2:k=31+d;break;  
        case 3:k=31+28+d;break;  
        case 4:k=31+28+31+d;break;  
        case 5:k=31+28+31+30+d;break;  
        case 6:k=31+28+31+30+31+d;break;  
        case 7:k=31+28+31+30+31+30+d;break;  
        case 8:k=31+28+31+30+31+30+31+d;break;  
        case 9:k=31+28+31+30+31+30+31+31+d;break;  
        case 10:k=31+28+31+30+31+30+31+31+30+d;break;  
        case 11:k=31+28+31+30+31+30+31+31+30+31+d;break;  
        case 12:k=31+28+31+30+31+30+31+31+30+31+30+d;break;  
    }  
        i=(y-1)/4;  
        j=(y-1)/400;  
        l=(y-1)/100;  
        n=(y-1)%4;  
   		r=(365+365+365+366)*i+n*365+k-l+j;  
        if(y%400==0&&m>2)    
        p=r+x+1;            
        else if(y%4==0&&m>2&&y%100!=0)   
        p=r+x+1;           
        else  
        p=r+x;
		int oneb,fourb;
		oneb=p/((365*3+366)*25);
		fourb=p/(((365*3+366)*25)*4);
		int p1;
		p1=p+oneb-fourb;
		int y1,m1,d1,y2,y3;
		y1=p1/(366+365*3);
		y2=(p1-y1*(366+365*3))/365;
		y3=4*y1+y2+1;
		d1=p1-y1*(366+365*3)-y2*365;
		if(d1<=31)
		m1=1;
		else if(d1<=31+28)
		m1=2;
		else if(d1<=31+28+31)
		m1=3;
		else if(d1<=31+28+31+30)
		m1=4;
		else if(d1<=31+28+31+30+31)
		m1=5;
		else if(d1<=31+28+31+30+31+30)
		m1=6;
		else if(d1<=31+28+31+30+31+30+31)
		m1=7;
		else if(d1<=31+28+31+30+31+30+31+31)
		m1=8;
		else if(d1<=31+28+31+30+31+30+31+31+30)
		m1=9;
		else if(d1<=31+28+31+30+31+30+31+31+30+31)
		m1=10;
		else if(d1<=31+28+31+30+31+30+31+31+30+31+31)
		m1=11;
		else
		m1=12;
		int d2=0;
		switch(m1)  
    {  
        case 1:d2=d1;break;  
        case 2:d2=d1-31;break;  
        case 3:d2=d1-(31+28);break;  
        case 4:d2=d1-(31+28+31);break;  
        case 5:d2=d1-(31+28+31+30);break;  
        case 6:d2=d1-(31+28+31+30+31);break;  
        case 7:d2=d1-(31+28+31+30+31+30);break;  
        case 8:d2=d1-(31+28+31+30+31+30+31);break;  
        case 9:d2=d1-(31+28+31+30+31+30+31+31);break;  
        case 10:d2=d1-(31+28+31+30+31+30+31+31+30);break;  
        case 11:d2=d1-(31+28+31+30+31+30+31+31+30+31);break;  
        case 12:d2=d1-(31+28+31+30+31+30+31+31+30+31+30);break;  
    }  
    if(y3%4==0&&y3%100!=0&&m1>2)
    {
		d2=d2-1;	
	}
	if(y3%400==0&&m1>2)
	{
		d2=d2-1;
	}
	if(m1==1&&d2==0)
	{
		m1=12;
		d2=31;
		y3=y3-1;
	}
	if(y3%4==0&&y3%100!=0&&m1==3&&d2==0)
	{
		m1=2;
		d2=29;
	}
	if(y3%400==0&&m1==3&&d2==0)
	{
		m1=2;
		d2=29;
	}
	printf("%d.%d.%d\n",y3,m1,d2);
	return 0;
}