#include<stdio.h>
#define MONTHS 12
int run(int month);
int ping(int month);
int main(void){
int year,month;   
printf("input year and month:\n");
scanf("%d %d",&year,&month);   
if(year%4==0 && year%100!=0 || year%400==0)
run(month);   
else   
ping(month);
}
/*闰年*/
int run(int month){   
int days[MONTHS]={31,28,31,30,31,30,31,31,30,31,30,31};
int ndays;    ndays=days[month-1];
printf("Month %d has %2d days.\n",month,ndays);
return ;
}
/*平年*/
int ping(int month){
int days[MONTHS]={31,29,31,30,31,30,31,31,30,31,30,31};
int ndays;    ndays=days[month-1];
printf("Month %d has %2d days.\n",month,ndays);   
return ;
}
有很多都是重复的，怎么改进更效率点，

int year_sumday(int year,int month,int day)
{
    int sum=0;
    int rui[12]={31,29,31,30,31,30,31,31,30,31,30,31};
    int ping[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    int ruiflag=0;
    if((year%4==0 &&year%100!=0) || year%400==0) ruiflag=1;
    for(int index=0;index<month-1;index++)
    {
        if(ruiflag==1) sum+=rui[index];else sum+=ping[index];
    }
    sum+=day;
    return sum;
}

以前写的  可以参考一下

路过

不好意思了 我求的是单个月份的天数，你那是之和啊

#include<stdio.h>
#define MONTHS 12
int date(int month);
int main(void)
{
  int year,month,days;   
  printf("input year and month:\n");
  scanf("%d %d",&year,&month);   
  days=date(month);
  if(year%4==0 && year%100!=0 && month==2 || year%400==0)
     printf("years %d month %d has %2d days",year,month,days+1); //如果是闰年，并且是2月份，天数+1
  else 
     printf("years %d month %d has %2d days",year,month,days); //否则按days数组的天数输出，因为只有闰年二月是个特例，单拿出来，其他月份可以一致
}
/*求月份天数*/
int date(int month)
{   
  int days[MONTHS]={31,28,31,30,31,30,31,31,30,31,30,31};
  int ndays;
  ndays=days[month-1];
  return ndays;
}

[ 本帖最后由 核晴核理 于 2013-2-5 09:57 编辑 ]

我是给你参考啊  求单个月份天数 不是比求和更容易一些吗