Problem 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

題意：兩個(gè)素?cái)?shù) a、b 之間的區(qū)間 (a,b]?稱謂非素?cái)?shù)區(qū)間，給出一個(gè)數(shù) n，求 n 所在的非素?cái)?shù)區(qū)間長(zhǎng)度。

思路：先打表求素?cái)?shù)，若 n 為素?cái)?shù)，則長(zhǎng)度為0，若 n 為合數(shù)，找出相鄰兩素?cái)?shù)，長(zhǎng)度為兩素?cái)?shù)差

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<map> #include<stack> #include<ctime> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 5000005 #define MOD 1e9+7 #define E 1e-6 #define LL long long using namespace std; LL prime[N],cnt; bool bprime[N]; void make_prime() {memset(bprime,true,sizeof(bprime));bprime[0]=false;bprime[1]=false;for(LL i=2;i<=N;i++){if(bprime[i]){prime[cnt++]=i;for(LL j=i*i;j<=N;j+=i)bprime[j]=false;}} } int main() {make_prime();LL n;while(scanf("%lld",&n)!=EOF&&n){if(bprime[n])printf("0\n");else{for(LL i=0;i<cnt;i++)if(prime[i]<n&&prime[i+1]>n)printf("%lld\n",prime[i+1]-prime[i]);}}return 0; }

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