3518 Prime Gap
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 6597 | Accepted: 3775 |
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
Source
//给你一个数,是素数的话输出0,否则输出 与之相邻的两个素数差,筛选法求素数
#include <iostream> #include <stdio.h> #include <algorithm> #include <queue> #include <cmath> #include <string.h> #define N 1299709 using namespace std; bool b[1300000]; bool is_prime(int n) { int i,m=sqrt(double(n)); for(i=2;i<=m;i++) if(n%i==0) return 0; return 1; } int main() { int j,i; b[1]=1; for(i=4;i<=N;i+=2) b[i]=1; for(i=3;i<=N;i+=2) if(is_prime(i)) { for(j=i+i;j<=N;j+=i) b[j]=1; } int k; while(scanf("%d",&k),k) { if(is_prime(k)) printf("0\n"); else { j=k+1; i=k-1; while(b[j]) j++; while(b[i]) i--; printf("%d\n",j-i); } } return 0; }