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
10
11
27
2
492170
0
Sample Output
4
0
6
0
114
这道题题意是给出一个数n，找一个小于它的最大的正数，以及一个大于它的最小的质数，可以以给出的数字为基准点，分别向前和向后拓展，遇到质数就停下来，并标记距离，代码如下
其中用到判断一个数是否为质数的函数
```#include<iostream>
#include<cmath>
using namespace std;
bool isprime(int n){
for(int i=2;i<=sqrt(n);i++){
if(n%i==0){
return false;
}
}
return true;
}
int main(){
int num;
while(cin>>num&&num!=0){
int ans=0;
if(isprime(num)){
cout<<ans<<endl;
}
else{
int temp1=num;
while(!isprime(temp1)){
temp1++;
}
int temp2=num;
while(!isprime(temp2)){
temp2--;
}
ans=temp1-temp2;
cout<<ans<<endl;
}
}
}