#include "stdio.h"
bool ifLeapYear(int year)//判断是不是云年
{
 if((year%400==0)||((year%4==0)&&(year%100!=0)))
 return true;
 else
 return false;
}
void main()
{
 int year,date,i;
    int month1[12]={31,29,31,30,31,30,31,31,30,31,30,31};
 int month2[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    bool flag;
    printf("请输入年份和天数\n");
    scanf("%d%d",&year,&date);
    if(date>366)
    {
     printf("输入错误！\n");
     return;
    }
 flag=ifLeapYear(year);
 if(flag)
  for(i=0;i<12;i++)
  {
      if(date<=month1[i]) break;
      date=date-month1[i];
  }
 else
  for(i=0;i<12;i++)
  {
      if(date<=month2[i]) break;
      date=date-month2[i];
  }
  printf("是%d月%d号",i+1,date);
}

  小弟的思路如上。。可换会指针就不懂了
 
 指针的要求
 void month_day (( int year,, int yearday,, int* pmonth, int * pday)
year和yearday 是输入的年份和天数
pmonth和pday是要输出的日期