[暑假集训--数论]poj3518 Prime Gap

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 + n is 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

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.

Output

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.

Sample Input

10
11
27
2
492170
0

Sample Output

4
0
6
0
114

给个x，如果x夹在两个质数a,b之间，求b-a，否则输出0

在筛法的时候预处理下距离就好

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #define LL long long
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n;
 bool mk[];
 int p[],len;
 int ans[];
 inline LL LLabs(LL a){return a<?-a:a;}
 inline void getp()
 {
     memset(ans,-,sizeof(ans));
     for (int i=;i<=;i++)
     {
         if (!mk[i])
         {
             p[++len]=i;
             ans[i]=;
             for (int j=*i;j<=;j+=i)mk[j]=;
         }else ans[i]=ans[i-]+;
     }
     for (int i=;i>=;i--)
     {
         if (!ans[i])continue;
         ans[i]=max(ans[i],ans[i+]);
     }
 }
 int main()
 {
     getp();
     ans[]=-;
     while (~scanf("%lld",&n)&&n)printf("%d\n",ans[n]?ans[n]+:);
 }

poj 3518