Prime Gap
Time Limit: 5000MSMemory Limit: 65536K
Total Submissions: 8499Accepted: 4983
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
题目大意：两个连续素数a和b之间的区间称为非素数区间（包括后边的素数b）。
给你一个数N，求N所在非素数区间的长度。
如素数 23~29 之间的非素数区间为24 25 26 27 28 +素数29。非素数区间长度为6(5+1)
给你一个数25 则25所在的非素数区间长度就为6
思路：
用筛法求素数打表，用PrimeNum存放所有素数。
若N为素数，则输出长度为0
若N为合数，则找出相邻的两个素数，输出长度为两素数的差
#include<stdio.h>
int Prime[1300000],PrimeNum[100010];
int IsPrime()
{
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 i = 2; i <= 1300000; i++)
{
if(Prime[i])
PrimeNum[num++] = i;
}
return num;
}
int main()
{
int n;
int num = IsPrime();
while(~scanf("%d",&n) && n)
{
if(Prime[n])
{
printf("0\n");
continue;
}
else
{
for(int i = 0; i < num; i++)
{
if(PrimeNum[i] < n && PrimeNum[i+1] > n)
printf("%d\n",PrimeNum[i+1]-PrimeNum[i]);
}
}
}
return 0;
}
