求个高手教下  万年历月历第一个星期前面的空格要怎么弄   （注：因为我课本才看到了for 麻烦弄点我能看得懂的谢谢)

网上搜的  不知道是否准确  自己验证

#include<stdio.h>
#include<stdlib.h>
int monthday[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
/*数组monthday[13]存放每个月的天数*/
char monthname[13][6]={"","JAN","FEB","MAR","APR","MAY","JUNE",
                                        "JULY","AUG","SEP","OCT","NOV","DEC"};
/*数组monthname[][]存放每个月的英文名称，用于打印*/
char weekday[8][6]={"SUN","MAN","TUE","WED","THU","FRI","SAT","SUM"};
/*数组weekday[][]存放一周的每一天的英文名称，用于打印*/
int month2[6][7];
int leapyear(int year)
/*函数1：判断year是否为闰年*/
{if((year%4==0)&&(year%100)!=0||(year%400)==0)
  return 1;
 else
  return 0;
}
void print2(int month,int t)
/*函数2：按月打印日历*/
{ int i,j;
  printf("***  %s  ***\n",monthname[month]);
  for(i=0;i<=6;i++)
     printf("%5s",weekday[i]);
  printf("\n");
  for(i=0;i<=t;i++)
   {for(j=0;j<=6;j++)
        if(month2[i][j]==0)
    printf("      ");
       else
     printf("%5d",month2[i][j]);
     printf("\n");
   }
}
void calendar(int year)
/*函数3：主体函数，按月生成日历*/
{int month;
 int todayweek,today,i,j,t;
 if(leapyear(year))
  monthday[2]=29;
 else
  monthday[2]=28;
 todayweek=year+(year-1)/4-(year-1)/100+(year-1)/400;
 todayweek=todayweek%7;
 /*计算当年第1天是星期几*/
 printf("===  year   %d  ===\n",year);
  for(month=1;month<=12;month++)
   {today=1;
    for(i=0;i<=5;i++)
       for(j=0;j<=5;j++)
        month2[i][j]=0;
     t=0;
     while(today<=monthday[month])
      {month2[t][todayweek]=today;
        todayweek++; today++;
         if(todayweek==7)
          {todayweek=0;
           t++;
          }
      }
    print2(month,t);
    }
}
main()
{int year;
 printf("请输入年份：\n");
 scanf("%d",&year);
 calendar(year);
system("pause");
}

学习

发现楼上的天天蹭分。。

此问题高贵不贵，简单实惠
2楼发的是正确的，复制下来只要改改就能运行，我正在研究它。涉及到C的很多东西，好好研究它有助于掌握知识，比我在网上找的那些题好太多了，主要是涉及面广，不再只是某一部分只是。
号召和我一样的新手一起研究

也来蹭个分

这样修改，就可以运行了
#include<stdio.h>
#include<stdlib.h>
int monthday[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
/*数组monthday[13]存放每个月的天数*/
char monthname[13][6]={"","JAN","FEB","MAR","APR","MAY","JUNE",
                                        "JULY","AUG","SEP","OCT","NOV","DEC"};
/*数组monthname[][]存放每个月的英文名称，用于打印*/
/*char weekday[8][6]={"SUN","MAN","TUE","WED","THU","FRI","SAT","SUＮ"};*/
char weekday[8][9]={"星期日","星期一","星期二","星期三","星期四","星期五","星期六","星期日"};
/*数组weekday[][]存放一周的每一天的英文名称，用于打印*/
int month2[6][7];

int leapyear(int year)
/*函数1：判断year是否为闰年*/
{if((year%4==0)&&(year%100)!=0||(year%400)==0)
  return 1;
else
  return 0;
}

void print2(int month,int t)
/*函数2：按月打印日历*/
{ int i,j;
  printf("***  %s  ***\n",monthname[month]);
  for(i=0;i<=6;i++)
     printf("%-8s",weekday[i]);
  printf("\n");
  for(i=0;i<=t;i++)
   {for(j=0;j<=6;j++)
        if(month2[i][j]==0)
            printf("%8s"," ");//printf("      ");
          else
             printf("%3s%-5d"," ",month2[i][j]);
        printf("\n");
   }
}
void calendar(int year)
/*函数3：主体函数，按月生成日历*/
{int month;
int todayweek,today,i,j,t;
if(leapyear(year))
  monthday[2]=29;
else
  monthday[2]=28;
todayweek=year+(year-1)/4-(year-1)/100+(year-1)/400;
todayweek=todayweek%7;
/*计算当年第1天是星期几*/
printf("===  year   %d  ===\n",year);
  for(month=1;month<=12;month++)
   {today=1;
    for(i=0;i<=5;i++)
       for(j=0;j<=6;j++)
            month2[i][j]=0;/*清空*/
     t=0;
     while(today<=monthday[month])
      {month2[t][todayweek]=today;
        todayweek++; today++;
         if(todayweek==7)
          {todayweek=0;
           t++;
          }
      }
    print2(month,t);
    }
}

main()
{int year;
printf("请输入年份：\n");
scanf("%d",&year);
printf("     一切有为法，如梦幻泡影，如露亦如电，应作如是观。\n\n");
calendar(year);
system("pause");
}

功夫不负苦心人，万年历研究成功，绝对完美

祝贺O(∩_∩)O哈哈~

printk(int k)
{
    int s;
    for(s=0;s<4*k;s++)
        printf(" ");
}
这里的k决定于当月的第一天星期几，而4*k，是因为我输出每个日期是%4d,希望能帮到你