Java 中的 Project Euler #50（解决方案无效）

Question

我的结果是错误的，但我找不到错误。

问题描述：

The prime 41, can be written as the sum of six consecutive primes:41 = 2 + 3 + 5 + 7 + 11 + 13. This is the longest sum of consecutive primes that adds to a prime below one-hundred. The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953. Which prime, below one-million, can be written as the sum of the most consecutive primes?

我的想法：

Starting from the first prime 2, compute the longest sum of consecutive primes that adds to a number below 1 million.

Count down from the longest, for each certain length, compute the sum of seri start from 2, then compute the sum of seri start from the second prime...

Stop when a sum is a prime.

我的代码：

public class Prob50 {
public static void main(String[] args) {
    // TODO Auto-generated method stub
    long sum=0;
    int count=0;
    for(int i = 2; ; i++){
        if(isPrime(i)){
            if((sum+=i)>1000000){
                sum-=i;
                break;
            }
            count++;
        }
    }

    int jump=0;
    boolean isOver= false;
    boolean isAns= false;
    for(;count>0;count--){
        jump=0;

        for(;;){
            int tempj=jump;
            int tempc=count;
            sum=0;
            for(int i = 2;tempc>0 ; i++){

                if(isPrime(i)&&tempj>0){
                    tempj--;
                    continue;
                }

                if(isPrime(i)){
                    tempc--;
                    if((sum+=i)>1000000){
                        sum-=i;
                        isOver=true;
                    }
                }
            }

            if(isPrime(sum)){
                isAns=true;
                break;
            }
            if(isOver){
                break;
            }
            jump++;
        }

        if(isAns){
            break;
        }

    }
    System.out.println(sum+" "+count);
}

private static boolean isPrime(long n){
    for(int i = 2 ; i <= Math.sqrt(n) ; i++){
        if(n%i==0){
            return false;
        }
    }
    return true;
}
}

我的结果：

958577 536

答案是 997651，计数应该是 543。

Answer 1

我想出来了，只需要补充一下：

isOver= false;

介于 isOver= false; 和 for(;;){ 之间。

Java 中的 Project Euler #50（解决方案无效）

Project Euler #50 in Java(Solution not working)

java

algorithm

debugging