System.out.print("*");
System.out.print(i);
System.out.print("=");
System.out.print(i * j);
System.out.print("\t");
}
System.out.println();
}
}
public static void main(String[] args) {
System.out.print("录入一个整数：");
Scanner input = new Scanner(System.in);
// 没有考虑判断输入的是否是整数。
int maxNum = input.nextInt();
new T(maxNum).print();
}
}
9. 程序设计（满分9分）
从键盘输入一个日期，格式为yyyy-M-d
要求计算该日期与1949年10月1日距离多少天
例如：
用户输入了：1949-10-2
程序输出：1
用户输入了：1949-11-1
程序输出：31
SimpleDateFormat sf = new SimpleDateFormat("yyyy-MM-dd");
Date d= new Date();
Date a = sf.parse("1949-10-11");
long x = d.getTime()-a.getTime();
System.out.println(x/1000/(24*60*60));
import java.util.Date;
import java.text.*;
import java.util.*;
public class time {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
SimpleDateFormat sf = new SimpleDateFormat("yyyy-MM-dd");
Date d= new Date();
Date a= new Date();
Date a = sf.parse("1949-10-11");
Scanner input = new Scanner(System.in);
long x = d.getTime()-a.getTime();
System.out.println(x/1000/(24*60*60));
}
}
是调不通的。求解。。。
zgh610815707
0位粉丝
17楼
引用 杨婷的问问(16楼)
如果是计算某一天距离1948-10-1日有多少天的话 也很简单 下面的代码：import java.util.Date; import java.text.*; import java.util.*; public class Date1{ public static void main(String[] args) { &n...
非常感谢、这个我们都没有学到呢、谢谢！！
#include
struct date {
int year;
int month;
int day;
};
int main (void) {
int getDaysOfDate (struct date d);
struct date firstDate,secondDate;
int firstDays,secondDays;
printf ("Enter first date [yyyy mm dd]:");
scanf ("%i %i %i",&firstDate.year,&firstDate.month,&firstDate.day);
printf ("Enter second date [yyyy mm dd]:");
scanf ("%i %i %i",&secondDate.year,&secondDate.month,&secondDate.day);
firstDays=getDaysOfDate(firstDate);
secondDays=getDaysOfDate(secondDate);
printf ("Two date between the number days is: %i\n",secondDays-firstDays
);
return 0;
}
int getDaysOfDate (struct date d) {
int days;
if(d.month<=2) {
days=1461*(d.year-1)/4+153*(d.month+13)/5+d.day;
}else{
days=1461*d.year/4+153*(d.month+1)/5+d.day;
}
if(d.year>=1700&&d.month>=3&&d.day>=1 && d.year<=1800&&d.month<=2&&d.day
<=28) {
days=days+2;
}
if(d.year>=1800&&d.month>=3&&d.day>=1 && d.year<=1900&&d.month<=2&&d.day
<=28) {
days=days+1;
}
return days;
}
public static int getDays(GregorianCalendar g1, GregorianCalendar g2) {
int elapsed = 0;
GregorianCalendar gc1, gc2;
if (g2.after(g1)) {
gc2 = (GregorianCalendar) g2.clone();
gc1 = (GregorianCalendar) g1.clone();
} else {
gc2 = (GregorianCalendar) g1.clone();
gc1 = (GregorianCalendar) g2.clone();
}
gc1.clear(Calendar.MILLISECOND);
gc1.clear(Calendar.SECOND);
gc1.clear(Calendar.MINUTE);
gc1.clear(Calendar.HOUR_OF_DAY);
gc2.clear(Calendar.MILLISECOND);
gc2.clear(Calendar.SECOND);
gc2.clear(Calendar.MINUTE);
gc2.clear(Calendar.HOUR_OF_DAY);
whil