int m,i,k,h=0,leap=1;
for(m=1;m<=101;m++)
{ k=Math.sqrt(m+1);
for(i=2;i<=k;i++)
if(m%i==0)
{leap=0;break;}
for(m=1;m<=101;m++)
{ k=Math.sqrt(m+1);
for(i=2;i<=k;i++)
if(m%i==0)
{leap=0;break;}
public class Priem
{
public static void main(String[] args)
{
boolen flag = true;
for (int i = 2;i < 100; i++)
{
flag = true;
for (int j = 2; j <= (int)Math.sqrt(i); j++)
{
if (i % j == 0)
{
flag = false;
break;
}
}
if (flag)
{
System.out.println("priem = " + i);
}
}
}
}
class Count {
public static void main(String[] args) {
long jiecheng = 1; //用于存每个数的阶乘
double sum = 0;//用于存阶乘的加和
for (int j = 1; j <= 10; j++) {
jiecheng *= j;
System.out.println(j + "的阶乘是:" + jiecheng);
sum += jiecheng;
}
System.out.println(sum);
}
}运行的结果:
1的阶乘是:12的阶乘是:23的阶乘是:64的阶乘是:245的阶乘是:1206的阶乘是:7207的阶乘是:50408的阶乘是:403209的阶乘是:36288010的阶乘是:3628800sum = 4037913.0-----------------------------这样比楼主写的简单不少:)
public class YinZi {
public static void main(String[] args){
//计算45,60的最大公因子和最小公倍数
int yinZi = 0;
for(int i =45/2;i>2;i--){
if (45%i==0&&60%i==0){
yinZi = i;
break;
}
}
System.out.println("最大公因子为:" + yinZi);
System.out.println("最小公倍数为:"+45/yinZi*60);
}
}运行结果:
最大公因子为:15最小公倍数为:180
public static void main(String []args)
{
int i = 60,j = 45;
int temp1,temp2;
temp1 = i;
temp2 = j;while (temp1 % temp2 != 0)
{
temp1 %= temp2;
if(temp1 < temp2)
{
temp1 ^= temp2;
temp2 ^= temp1;
temp1 ^= temp2;
}
}
System.out.println("最大公约:"+temp2);
System.out.println("最小公倍:"+i*j/temp2);
}