要發就發

/*
 * 假設今年是1993年，現在將不超過1993的所有素數從小到大排成第一行，第二行上的每個數都等於它右肩上素數與其左肩上素數之差。問：第二行樹種是否存在若干連續的整數，他們的和恰好是1898。假如存在的話，又有幾種這樣的情況？
 * StoryMonster   2016/7/12
 */
#include <iostream>
#include <vector>

using namespace std;

bool PrimeNumber(int n)
{
    if(n==0 || n==1) return false;
    for(int i=2;i<n/2;i++)
    {
        if(n%i == 0) return false;
    }
    return true;
}
int main()
{
    vector<int> prime;
    vector<int> cha;
    prime.push_back(2);
    int counter = 0;
    for(int i=3;i<=1993;i++)
    {
        if(PrimeNumber(i))
        {
            counter++;
            prime.push_back(i);
            vector<int>::iterator p = prime.begin()+counter-1;
            cha.push_back(i-*p);
        }
    }
    cout << "素數:"<<endl;
    for(vector<int>::iterator prime_p = prime.begin();prime_p!=prime.end();prime_p++)
    {
        cout<<*prime_p<<" ";
    }
    cout<<endl;
    cout << "素數差:"<<endl;
    for(vector<int>::iterator cha_p = cha.begin();cha_p!=cha.end();cha_p++)
    {
        cout << " "<<*cha_p;
    }
    cout<<endl;
    int count = 0;
    vector<int>::iterator err_ptr1 = cha.begin();
    vector<int>::iterator err_ptr2 = cha.begin();
    vector<int>::iterator prime_ptr1 = prime.begin();
    vector<int>::iterator prime_ptr2 = prime.begin();
    while(err_ptr2!=cha.end())
    {
        count += *err_ptr2;
        if(count == 1898)
        {
            cout << "起始"<<*prime_ptr1<<"  終止"<<*(prime_ptr2+1)<<endl;
            count -= *err_ptr1;
            err_ptr1++;
            err_ptr2++;
            prime_ptr1++;
            prime_ptr2++;
        }
        else if(count > 1898)
        {
            count -= *err_ptr1;
            err_ptr1++;
            prime_ptr1++;
            count -= *err_ptr2;
        }
        else 
        {
            err_ptr2++;
            prime_ptr2++;
        }
    }
    return 0;
}