sicily 1500. Prime Gap

1500. Prime Gap

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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

题目分析

如果输入为素数,则输出0
否则输出左右两个素数的差
考察筛法求素数

#include <stdio.h>
#include <memory.h>

int prime[100000];
int count;

void init() {
  bool visited[1299710];
  memset(visited, true, sizeof(visited));
  for (int c = 2; c < 1299710; ++c)
    if (visited[c])
      for (int d = c + c; d < 1299710; d += c)
        visited[d] = false;
  count = 0;
  for(int c = 2; c < 1299710; ++c)
    if (visited[c])
      prime[count++] = c;
}

int main()
{
  init();
  
  int key;
  while (scanf("%d", &key) && key) {
    for (int c = 0; c < count; ++c) {
      if (key == prime[c]) {
        printf("0\n");
        break;
      } else if (key < prime[c]) {
        printf("%d\n", prime[c] - prime[c - 1]);
        break;
      }
    }
  }
  return 0;
}