import java.util.Scanner;
class JiSuanTianShu//计算年份1到年份2之间的天数
{
    public static void main (String [] args){
    Scanner s = new Scanner(System.in);
    System.out.print("请输入起始年：");
    int y1 = s.nextInt();
    System.out.print("请输入始年月：");
    int m1 = s.nextInt();
    System.out.print("请输入始年日：");
    int d1 = s.nextInt();
    System.out.print("请输入截止年：");
    int y2 = s.nextInt();
    System.out.print("请输入截止月：");
    int m2 = s.nextInt();
    System.out.print("请输入截止日：");
    int d2 = s.nextInt();
    int sum = 0 ;
    int sum1 = 0;
    int sum2 = 0;
    int sum3 = 0;
    if(y1<y2)//判断起止年
   {  
//计算起止年之间整年的天数   
      for(int i= y1+1 ; i<y2 ;i++){
          if(i%4==0&&i%100==00||i%400==0)
           { sum3 = sum3 + 366;
          System.out.println(i+"年"+sum3+"天");}
          else
            {sum3 = sum3 + 365;
            System.out.println(i+"年"+sum3+"天");}
       }
//计算起始年剩余的天数。
       switch(m1){
        case 1:
           sum1 = sum1+31;
        case 2:
          if(y1%4==0&&y1%100!=0||y1%400==0)
           sum1 = sum1+29;
          else
           sum1 = sum1+28;
        case 3:
           sum1 = sum1+31;
        case  4:
           sum1 = sum1+30;
        case  5:
           sum1 = sum1+31;
        case 6 :
           sum1 = sum1+30;      
        case 7:
           sum1 = sum1+31;
        case  8:
           sum1 = sum1+31;   
        case 9:
           sum1 = sum1+30;     
        case 10:
           sum1 = sum1+31;
        case 11:
           sum1 = sum1+30;
        case 12 :
           sum1 = sum1+31;
      
       }
         sum1 = sum1-d1;  
         System.out.println(y1+"年"+m1+"月"+d1+"日"+"之后"+y1+"年还剩余"+sum1+"天");
//计算截止年月日之前的天数。
       switch(m2-1){
         case 11:
           sum2 = sum2+30;
         case 10:
           sum2 = sum2+31;
         case 9:
           sum2 = sum2+30;
         case  8:
           sum2 = sum2+31;
         case 7:
           sum2 = sum2+31;
         case 6 :
           sum2 = sum2+30;
         case  5:
           sum2 = sum2+31;
         case  4:
           sum2 = sum2+30;
         case 3:
           sum2 = sum2+31;
         case 2:
           { if(y1%4==0&&y1%100!=0||y1%400==0)
               sum2 = sum2+29;
             else
               sum2 = sum2+28;
            }
         case 1 :
           sum2 = sum2+31;
        }
           sum2 = sum2+d2;
            System.out.println(y2+"年"+m2+"月"+d2+"日是"+y2+"年的第"+sum2+"天。");
     sum = sum1 +sum2+sum3;
     System.out.println(y1+"年"+m1+"月"+d1+"日"+"到"+y2+"年"+m2+"月"+d2+"日有"+sum+"天。");
    }else if(y1==y2)//同年则要判断是否月份大小以及日期大小
       {
          if(m1<=m2&&d1<=d2)
            {
              sum = sum1+d2-d1;     
              System.out.println(y1+"年"+m1+"月"+d1+"日"+"到"+y2+"年"+m2+"月"+d2+"日有"+sum+"天。");
           }else  if(d1>d2){
              System.out.println("起止日期输入错误！");

           }else
              System.out.println("起止月份输入错误！");

       }else
          System.out.println("起止年份输入错误！");
}
}

总感觉计算机的执行顺序有问题，尤其是年份相等分支时。

计算没这么麻烦吧

请教如何优化？？？？
请指点一下！谢谢！！

c语言里有个关于time 的 tm结构体  不知道java里是否有

不用time函数之类，纯循环和分支结构编写。

这样写看起来有些乱  适当的使用函数 会好一些   比如 计算某年是365天还是366天  计算X年x月x日是一年的第几天  分别单独写一个函数

把各月天数放进数组，平年闰年另外计算

这么简单的问题，何必把它复杂化呢？
SimpleDateFormat sdf = new SimpleDateFormat("yyyy年MM月dd日");
int year1 = 2011, month1 = 1, date1 = 23;
int year2 = 2016, month2 = 1, date2 = 26;
        
Date d = new Date(year1 - 1900, month1 - 1, date1) {};
Date d2 = new Date(year2 - 1900, month2 - 1, date2) {};
int days = (int)((d2.getTime() - d.getTime()) / 3600000 / 24);
System.out.println(sdf.format(d)+" 到 " + sdf.format(d2) +" 相差 " + days+ " 天.");

[此贴子已经被作者于2016-3-2 11:59编辑过]