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开1229大的数组,附代码

Posted by fym0512 at 2011-03-25 17:09:47 on Problem 2739
// Sum of Consecutive Prime Numbers.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<iostream>
#include<iomanip>
#include<cmath>
#include<fstream>
using namespace std;

int main(int argc, char* argv[])
{
	ifstream cin("a.txt");
	int n,summ;
	int prim[1229];
	int i,j,sum=0;
	double m;
	//将1-10000之间的素数保存到prim数组中
	for(m=2;m<=10000;m++)
	{
		for(j=2;j<=sqrt(m);j++)
			if((int)m%j==0)
				goto here;
			prim[sum++]=m;
here:;
	}
/*	cout<<"sum="<<sum<<endl;
	for(i=0;i<sum;i++)
		printf("%10d",prim[i]);
*/
	while(cin>>n)
	{
		if(n==0)
			break;
		summ=0;
		
	for(i=0;i<sum;i++)
	{
		int su=0;
		for(j=i;;j++)
		{
			su+=prim[j];
			if(su>n)
				break;
			if(su==n)
				
			{
	//			for(int k=i;k<=j;k++)
	//		cout<<setw(7)<<prim[k];
	//		cout<<endl;
				summ++;
			}
				
		}
	}
	
/*	for(i=2;i<sqrt(n);i++)
		if(n%i==0)
		{break;summ--;}
		summ++;
*/
	printf("%d\n",summ);

	}

	return 0;
}

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