Problem 50 Consecutive prime sum （线性筛）

Consecutive prime sum

Problem 50

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?

Answer:

997651Completed on Mon, 31 Oct 2016, 10:49

代码：

#include<bits/stdc++.h>using namespace std;vector<int> primes;int prime[1000000];void init(){    for(int i = 0 ; i < 1000000 ; i++){        prime[i] = 1;       }    for(int i = 2 ; i < 1000000; i++){        if (prime[i]){            primes.push_back(i);            if(i <= (int)sqrt((double)1000000)){                    for(int t = i*i ; t<1000000 ; t+=i)//prime[i*i+k*i]=false{                       prime[t] = 0;                }            }        }    }}int main(){        init();    int max = 0;    int maxlen = 0, maxsum = 0, cursum = 0;    for(int i = 2 ; i < primes.size() ; i++){        cursum = 0;        for(int j = i ; j < primes.size() ; j++)        {            cursum+=primes[j];            if (cursum>=1000000)  break;            if (prime[cursum] && j - i > maxlen )            {                maxlen = j-i;                maxsum = cursum;            }        }    }    cout<<maxsum<<endl;;    return 0;}