C语言程序设计：给出年、月、日，计算该日是该年的第几天。

Question

如题所述

Answer 1

#include "stdio.h" 

int main () {

  int year;

  int month;

  int day;

  scanf("%d %d %d", &year, &month, &day);

  if(year < 1 || (month < 1 || month > 12) || (day < 1 || day > 31))return -1;

  int daynum[12] = {

    31, 28, 31,30 ,31, 30, 31, 31, 30, 31, 30, 31

  };

  if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) {

    daynum[1] = 29;

  }

  int dayth = 0;

  for (int i = 0; i < month - 1; ++i) {

    dayth += daynum[i];

  }

  dayth += day;

  printf("the day is:%d\n", dayth);

  return 0;

}

Answer 2

#include <stdio.h>
int day_of_year(int year,int month, int day)
{
int monthDays[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int i;
for(i=0;i<month;i++)
day+=monthDays[i];
if(year%4==0&&year%100!=0||year%400==0)
{
if (month >2)
day+=1;
}
return day;
}
int main()
{
int day;
day=day_of_year(2013,5,23);
printf("%d\n",day);
return 0;
}

Answer 3

任意输入a年b月c日，天数用n表示
b<2,n=31+c(最多)
b>3,要判断a年是否为闰年，if(a/400==0||(a/4==0&&a/100!=0))就是闰年
n=31+....追问

有程序语言吗

Answer 4

#include<stdio.h>
int main()
{
int n,y,r,t,i;
int a[12];
a[1]=31; a[2]=28; a[3]=31;
a[4]=30; a[5]=31; a[6]=30;
a[7]=31; a[8]=31; a[9]=30;
a[10]=31;a[11]=30;a[12]=31;
while (scanf("%d %d %d",&n,&y,&r)!=EOF)
{
t=0;
if((n%4==0 && n%100!=0)||(n%400==0)) a[2]=29; else a[2]=28;
for (i=1;i<y;i++)
t+=a[i];
t+=r;
printf("%d\n",t);
}
}

Answer 5

#include<stdio.h>
struct
{
int year;
int month;
int day;
}date;
void main()
{
int i,days=0;
int day_month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
printf("input year,mont,day:\n');
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
for(i=0;i<date.month;i++)
days=days+day_month[i];
days=days+date.day;
printf("%d年%d月%d日是该年的第%d天",date.year,date.month,date.day,days);
}