算法分析： 题意： 水题一枚：题意找到包含所给数的一个数列，该数列满足在两个相邻质数之间，数列包含最大的质数，举个例子，10，则7，8，9，10，11，则数列为8，9，10，11 分析： 直接打表欧克 代码实现 |
Prime Gap Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 10711 Accepted: 6163 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 Japan 2007 |
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
const long N = 1399709;
int cnt[N];
long prime[N] = {0},num_prime = 0;
int isNotPrime[N] = {1, 1};
int Prime()
{
for(long i = 2 ; i < N ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
//关键处1
for(long j = 0 ; j < num_prime && i * prime[j] < N ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j] ) ) //关键处2
break;
}
}
return 0;
}
int main()
{
Prime();
int n;
while(scanf("%d",&n)!=EOF&&n!=0)
{
int ans=0;
for(int i=0;i<=1299709;i++)
{
if(prime[i]>n)
{
ans=prime[i]-prime[i-1];
break;
}
if(prime[i]==n)
{
ans=0;break;
}
}
printf("%d\n",ans);
}
return 0;
}
