Question

”输入某年某月某日，判断这一天是这一年的第几天“用C语言怎样编程

一个C语言编程题目“输入某年某月某日，判断这一天是这一年的第几天？”怎样用C语言编程

Answer 1

#include<stdio.h>
void main()
{
int sumday(int month,int day);/*计算第多少天的函数声明*/
int leap(int year);/*判断是否闰年的函数声明*/
int month,day,days,year;
int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};/*一般年的12个月的天数组成一个数组*/
printf("Please input date(year,month,day):");/*提示输入年月日*/
scanf("%d-%d-%d",&year,&month,&day);/*输入年月日*/
printf("%d/%d/%d\n",year,month,day);
days=sumday(month,day);/*计算天数的函数调用*/
if(leap(year)&&month>=3)/*调用判断是否闰年的函数，并且月份大于等于3时天数加1*/
days=days+1;
printf("Today is the %d day of this year.",days);/*显示是第多少天的函数*/
}
int sumday(int month,int day)
{
int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int i;
for(
i=0;i<month;i++)
day+=daytab[i];
return(day);
}

int leap(int year)
{
int leap;
leap=(year%4==0&&year%100!=0||year%400==0);
return(leap);
}

Answer 2

main()
{
int day,month,year,sum,leap;
printf("\nplease input year,month,day\n");
scanf("%d,%d,%d",&year,&month,&day);
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("It is the %dth day.",sum);}