public class Test2 {
public static void main(String[] args) {
int n = 101;
int m = 1000;
for(int i=2; i<Math.sqrt(n); i++){
if(n % i != 0){
System.out.print(n+" ");
++n;
if(n>m) {break;}
}
}
}
}
为什么我怎么改M输出的结果都是 101 102 103 104 105 106 107
public static void main(String[] args) {
int n = 101;
int m = 1000;
for(int i=2; i<Math.sqrt(n); i++){
if(n % i != 0){
System.out.print(n+" ");
++n;
if(n>m) {break;}
}
}
}
}
为什么我怎么改M输出的结果都是 101 102 103 104 105 106 107
public static void main(String[] args) {
for(int i=100;i<=200;i++){
for(int j=2;j<i/2;j++){
if(i%j==0)
break;
else{
System.out.println(i);
break;
}
}
}
}
}
i的值:2, 3, 4, 5, 6, 7, 8, 10
i = 2时,n=101,n%2 != 0, 打印101,++n, n = 102
i = 3时,n=102,n%3 = 0,
i = 4时,n=102,n%4 != 0, 打印102,++n, n = 103
i = 5时,n=103,n%5 !=0, ,打印103,++n, n=104
i = 6时,n=104, n%6 != 0, 打印104,++n, n=105
i = 7时,n=105, n%7 != 0, 打印105,++n, n=106
i = 8时,n=106, n%8 != 0, 打印106,++n, n=107
i = 9时,n=107, n%9 = 0
i = 10时,n=107, n%10 != 0, 打印107,++n, n=108
.....其他LZ自己考虑吧至于求101到200之间的质数,可以这样:public class Test {
public static void main(String[] args) {
for(int n=101; n<=200; n++) {
boolean flag = true;
for(int i=2; i<Math.sqrt(n)+1; i++) {
if(n % i == 0) {
flag = false;
break;
}
}
if(flag == true)
System.out.print(n + " ");
}
System.out.println();
}
}
public class Prime {
// 判断num是不是素数,是则返回true
public static boolean isPrime(int num) {
if (num == 2 || num == 3)
return true;
if (num % 2 == 0)// num为偶数
return false; // num为奇数
int d = 3;
while (d <= Math.sqrt(num) && num % d != 0) {
d += 2;
}
if (num % d != 0)
return true;
return false;
} public static void main(String[] args) {
for (int i = 101; i < 200; ++i) {
if (isPrime(i))
System.out.println(i);
}
}}