用java编写一个输出1到00以内的素数并求和的程序

用java编写一个输出1到00以内的素数并求和的程序

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

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。

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; } }

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; } } }

使用你上面的那段判断代码： int sum=0; for(int i=100;i<=1000;i++){ if(isprime(i){ sum+=i; } } system.out.println(sum);

public static int getPrimeSumOf(int num ){ int sum = 0; boolean [] primes = primeNumbers(num); StringBuilder sb = new StringBuilder(); if(primes != null){ for(int i = 1; i< primes.length; i++){ if(primes[i]){ // System.out.print(i+" + ");//输出 sb.append(i+" + "); sum += i; //求和 } } System.out.println(sb.toString().substring(0, sb.toString().length()-2)); System.out.println("= "+ sum); } return sum; } public static boolean[] primeNumbers(int num){ if(num <=0){ System.out.println("范围必须大于0"); return null; } boolean [] primes = new boolean[num+1]; primes[1] = false; //排除1 Arrays.fill(primes, 2, num+1, true);//从2开始，假设都为素数 int n = (int) Math.sqrt(num); for(int i = 1; i< n; i++){ if(primes[i]){//如果i是质数，那么i的倍数不是质数 for(int j = 2*i; j<=num; j+=i){ primes[j] = false; } } } return primes; } ---------------------------------- public static void main(String[] args) { getPrimeSumOf(100); }