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 numbersp 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 containsk.

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

10112724921700

Sample Output

4060114

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;}