POJ3518 Prime Gap - 水题

题目连接

http://poj.org/problem?id=3518

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 + 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

题意

定义素数区间的长度为相邻两个素数的差,求一个数所在素数区间的长度。

题解

用之前的打表模板会MLE,所以换了个模板,打表后输出即可。

代码

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#include <iostream>
using namespace std;
const int maxn = 1000010,maxnn=2000000;
int prime[maxn];
bool vis[maxnn],is_prime[maxnn];
void init_prime_factor(int n)
{
int num=0;
memset(vis,true,sizeof(vis));
vis[0]=vis[1]=false;
for(int i=2;i<n;i++)
{
if(vis[i]) {
prime[num++]=i;
is_prime[i]=true;
}
for(int j=0;j<num && i*prime[j]<n;j++)
{
vis[i*prime[j]]=false;
if(i%prime[j]==0) break;
}
}
}
int main(){
int n;
init_prime_factor(maxnn);
while(~scanf("%d",&n)){
if(n==0) break;
if(is_prime[n] || n==1){
printf("0\n");
continue;
}
for(int i=0;i<maxn;++i){
if(prime[i]>n){
printf("%d\n",prime[i]-prime[i-1]);
break;
}
}
}
}