第1个回答 2012-03-20
提供个法子给你参考,不是效率最好的。
iResult 是求和结果,sResult 是输出结果,你在后边加输出语句输出即可。
for (int i = 2; i <= 100; i++) {
for (int j = 1; j <= i; j++) {
if (j != 1 && j != i && i%j == 0) {
break;
}
if (j == i) {
if (i == 2) {
sResult = sResult + i;
} else {
sResult = sResult + "," + i;
}
iResult = iResult + i;
}
}
}
第2个回答 2012-03-20
public class Prime {
static boolean isPrime(int k){
if(k==2)
return false;
if(k%2==0)
return false;
int j=(int)Math.sqrt(k);
if(j%2==0)
j--;
while(j>2&&k%j!=0)
j-=2;
return j<2;
}
public static void main(String args[]){
int sum=2;
System.out.println("2 ");
for(int k=3;k<=100;k+=2){
if(isPrime(k)){
sum=sum+k;
System.out.println(k);
}
}
System.out.println("素数的和为"+sum);
}
}
主要是判断素数,这个方法有很多。
第3个回答 2012-03-20
public class Test {
public static void main(String[] args) {
int sum = 0;
int i = 2;
while(i <= 100){
boolean isPrime = true;
for(int j = 2; j < i; j++){
if(i % j == 0){
isPrime = false;
break;
}
}
if(isPrime){
sum = sum + i;
}
i++;
}
System.out.println("Sum of prime number between 1 and 100 is: " + sum);
}
}
----------testing
Sum of prime number between 1 and 100 is: 1060本回答被提问者采纳
第4个回答 2012-03-20
public class BT {
public static void main(String[] args) {
int sum = 0;
System.out.print("2 + 3");
sum += 2 + 3;
for (int i = 4; i < 100; i++) {
if (isCable(i)) {
System.out.print(" + " + i);
sum += i;
}
}
System.out.print(" = " + sum);
}
public static boolean isCable(int number) {
for (int i = 2; i <= number / 2; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}
}