• UVA


    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 gapof length n . For example, $ langle$24, 25, 26, 27, 28$ 
angle$ 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
    
    
    
    

    为了完成规定的题数,我只能先找找水题做做了。。。



    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1400000;
    bool isp[maxn];
    int res[maxn];
    
    void init()
    {
        isp[0] = isp[1] = false;
        for(int i = 2; i < maxn; i++)
            isp[i] = true;
        int Max = sqrt(maxn) + 1;
        for(int i = 2; i < Max; i++)
            if(isp[i])
                for(int j = 2; i*j < maxn; j++)
                    isp[i*j] = false;
        int l, r;
        for(int i = 2; i < maxn; i++){
            if(isp[i]) l = i, res[i] = 0;
            else res[i] = l;
        }
        for(int i = 1299709; i >= 2; --i){
            if(isp[i]) r = i;
            else res[i] = r - res[i];
        }
    }
    
    int main()
    {
        init();
        int n;
        while(cin >> n, n!=0)
            cout << res[n] << endl;
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/kunsoft/p/5312766.html
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