集训第十二天（2017/8/11）：刷题（二分法）

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今天还是刷二分法的题，又做了一道莫名其妙的题，用scanf输入的话能ac，用cin输入就runtime error，我真的不知道这是什么情况？？贴上这道题（通过这道题也收获了很多，比如说筛法求素数这道题，也是一道挺好的题），有没有大佬知道怎么解决？

Prime Gap

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 30 Accepted Submission(s) : 24

Problem Description

The sequence of n 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + nis called a prime gap of length n. For example, 24, 25, 26, 27, 28 between 23 and 29 is a prime gap of length 6.

Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.

Input

<p>The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.</p>

Output

<p>The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.</p>

Sample Input

10112724921700

Sample Output

#include<iostream>
#include<cstdio>
using namespace std;
int prime[1300000],primenum[100010];
int isprime()//筛法求素数
{
    for(int i=2;i<=1300000;i++)
         prime[i]=1;
    for(int i=2;i<=1300000;i++)
      for(int j=i+i;j<=1300000;j+=i)
         prime[j]=0;
    int num=0;
    for(int k=2;k<=1300000;k++)
      {
          if(prime[k]) primenum[num++]=k;
      }
    return num;
}
int main()
{
    int num=isprime();
    int n;
    while(scanf("%d",&n)!=EOF && n)//while(cin>>n&&n)运行时错误？！
    {
        if(prime[n])
        {
            cout<<0<<endl;
            continue;
        }
        else
         for(int i=0;i<num;i++)
        {
            if(primenum[i]<n&&primenum[i+1]>n)
              cout<<(primenum[i+1]-primenum[i])<<endl;
        }
    }
    return 0;
}

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