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<>第一种方式
质数：素数
质数：只能被1和他本身整除的数字
最小的质数：2 3 5 7 11 13 17 19 23…
思路1
先得到2-100之间所有的数字
然后得到这个数字x需要%的数字[2到x-1]
如果2到x-1之间有一个数字可以被x整除的话
那么就计数器+1[计数器在统计除了1和他本身之外有几个数字可以被x整除]
等2到x-1之间所有的数字都判断完之后 看看计数器是不是为0
如果是0说明在这个区间里面没有数字可以被x整除
那么打印出来就可以
for(int x = 2;x <= 100;x++){//1-100之间所有的数字 //每一个数字都需要有单独的计数器 int count = 0;
//%其他的数字 x%1 == 0 x%x==0 x%比x大的数字=x for(int y = 2;y < x;y++){//其他的数字 if(x % y ==
0){ count++; } } if(count == 0){ System.out.println(x); } }
<>第二种方式
先得到2-100之间所有的数字
再得到这个数字x需要%的数字[2到x-1]
如何发现2到x-1之间有一个数字可以被x整除
说明x不是质数 那么应该跳过 看下一个数字x
比如x = 9 本来应该拿着9%[2-8]
但是当我们拿着9%3就已经和0相等 说明除了1和9之外
还有一个数字3可以被9整除 说明9不是质数
那么就没有必要拿着9继续%[4-8] 直接换成10这个数字就可以
a:for(int x = 2;x <= 100;x++){ for(int y = 2;y < x;y++){ if(x % y == 0){
continue a;//本来是结束本次循环,这里加了标签,直接结束本次最外边的for循环 } }//该%的数字都已经%完 System.out.println
(x); }
<>第三种方式
int count = 0; for (int x = 2; x < 100; x++) { boolean flag = true; for (int y
= 2; y < x;y++) { if (x % y == 0) {//如果可以除尽,说明不是素数,我们把标识换成false,这样就不会打印 flag =
false; break; } } if (flag) { System.out.println(x); count++; } }
