<分区>
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。