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work3

/用“埃氏筛法”求2~100以内的素数。2~100以内的数,
先去掉2的倍数,再去掉3的倍数,再去掉5的倍数,……依
此类推,最后剩下的就是素数。
/

work2

package test;
class Test{
public static void main(String[] args) {
int i,j,in,jn,sumi,a;
double sumj=0,ai;
int n;
for(j=1;j<=2000;j++) {
sumi=0;
for(i=j;i<=2000;i++) {
in=(int)Math.pow(i, 3);
sumi=sumi+in;
for(n=i;n<=2000;n++)
if((sumi==nnn)&&(n!=i)) {
for(a=j;a<=i;a++)
System.out.print(a+"^3+");
System.out.println("="+n+"^3");
}
}
}
}
}

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