2回答

• 

  慕寒州

  2017-07-31 07:08:55

  已采纳

  #include <stdio.h>
  int main() 
  { 
      /* 定义需要计算的日期 */
      int year = 2008;
      int month = 8;
      int day = 8;
      int flag,sum,sum1;
      switch(month){
          case 1:sum1=0;break;
          case 2:sum1=31;break;
          case 3:sum1=59;break;
          case 4:sum1=90;break;
          case 5:sum1=120;break;
          case 6:sum1=151;break;
          case 7:sum1=181;break;
          case 8:sum1=212;break;
          case 9:sum1=243;break;
          case 10:sum1=273;break;
          case 11:sum1=304;break;
          case 12:sum1=334;break;
      }
      if(year%4==0&&year%100)
          flag=1;
      else flag=0;
      if(month>=3) sum=sum1+flag+day;
      else sum=sum1+day;
      printf("%d年%d月%d日是该年的第%d天",year,month,day,sum);
  	return 0;
  }

  
  

  1 0

  慕工程653...

  厉害?

  2017-07-31 15:36:16

  共 1 条回复 >

• 

  慕工程6531016

  2017-07-31 16:52:46

  //我自己又写了一个给你们参考
  #include <stdio.h>

  int main()

  {

  int scor=2008;
  int q=8;
  int w=8;
  int a,s,d,f;
  a=0;
  if(scor%4==0)
  {
  d=29;
  f=366;
  }
  else
  {
  d=28;
  f=365;
  }
  switch(q)
  {
  case 1: a+=31;
  case 2: a+=d;
  case 3: a+=31;
  case 4: a+=30;
  case 5: a+=31;
  case 6: a+=30;
  case 7: a+=31;
  case 8: a+=31;
  case 9: a+=30;
  case 10: a+=31;
  case 11: a+=30;
  case 12: a+=31;
  }
  s=f-a+w;
  printf("%d\n",s);
  return 0;
  }

  0 1