第1个回答 2010-08-04
#include<stdio.h>
int days[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
void main()
{
int sum=0,a,b,c,i;
char ags='y';
while(ags=='y'||ags=='Y')
{
sum = 0;
printf("请输入年/月/日:");
scanf("%d%d%d",&a,&b,&c);
for(b--;b>0;b--)
{
i = days[b];
if((a%4==0&&a%100!=0)||(a%400==0)) i++;
sum+=i;
}
sum+=c;
printf("是这一年的%d天\n",sum);
printf("是否继续输入日期(y/n)?");
fflush(stdin);
ags=getchar();
}
printf("结束输入!\n");
}
------------
运行结果:
请输入年/月/日:2006 3 11
是这一年的70天
是否继续输入日期(y/n)?n
第2个回答 2010-08-04
#include<stdio.h>
void main()
{
int month[12]={30,28,31,30,31,30,31,31,30,31,30,31};
int days,i,j;
int mon,day,sum=0,year;
printf("输入日期,XXXX年XX月XX日\n");
scanf("%d%d%d",&year,&mon,&day);
i=mon;
for(j=0;j<mon-1;j++)
{
sum+=month[i];
}
if((year%4==0&&year%100!=0)||year%400==0)
sum=sum+1;
days=sum+day;
printf("%d",days);
}