POJ 3518 Prime Gap题意好难理解



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

题意：读了半天我也没读懂这题是个啥意思，看了别人的解释，果断AC，给你一个数n，如果是素数直接输出0，否则找起左右最近的素数求差值

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int ok(int x){
    for(int i=2;i<=sqrt(double(x));i++){
        if(x%i==0){
            return 0;
        }
    }
    return 1;
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF&&n){
        if(ok(n)){
            printf("0\n");
            continue;
        }
        int i,j;
        for(i=n;i>1;i--){
            if(ok(i)){break;}
        }
        for(j=n;;j++){
            if(ok(j)){break;}
        }
        printf("%d\n",j-i);
    }
    return 0;
}

注意：sqrt（int）c++编译器编译不过

我很无语，读不懂题简直就是白痴了。。。。。。。疯了