[sicily]1500. Prime Gap
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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 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
10112724921700
Sample Output
4060114
Problem Source
Tokyo 2007
简单计数题,要我们计算包含k值的相邻两个素数之间的距离。利用筛法求出N以内的素数,然后计算即可。代码如下:
#include <iostream>#include <cstring>#include <cmath>using namespace std;#define N 1299710bool prime[N];int main(){ int n; memset(prime, true, sizeof(prime)); //筛法求N以内的素数。 for(int i=2; i<=sqrt(N); i++) if(prime[i]) for(int j=i+i; j<N; j=i+j) prime[j] = false; while(cin>>n && n>0) { int count=0; if(!prime[n]) { int tmp=n; while(prime[tmp]==false) tmp--; tmp++; count++; while(prime[tmp]==false){ tmp++; count++; } } cout<<count<<endl; } // system("pause"); return 0; }
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