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  • CoderForces999B- Reversing Encryption

    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    A string ss of length nn can be encrypted by the following algorithm:

    • iterate over all divisors of nn in decreasing order (i.e. from nn to 11),
    • for each divisor dd, reverse the substring s[1d]s[1…d] (i.e. the substring which starts at position 11 and ends at position dd).

    For example, the above algorithm applied to the string ss="codeforces" leads to the following changes: "codeforces"secrofedoc "orcesfedoc "rocesfedoc "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1d=1).

    You are given the encrypted string tt. Your task is to decrypt this string, i.e., to find a string ss such that the above algorithm results in string tt. It can be proven that this string ss always exists and is unique.

    Input

    The first line of input consists of a single integer nn (1n1001≤n≤100) — the length of the string tt. The second line of input consists of the string tt. The length of tt is nn, and it consists only of lowercase Latin letters.

    Output

    Print a string ss such that the above algorithm results in tt.

    Examples
    input
    Copy
    10
    rocesfedoc
    
    output
    Copy
    codeforces
    
    input
    Copy
    16
    plmaetwoxesisiht
    
    output
    Copy
    thisisexampletwo
    
    input
    Copy
    1
    z
    
    output
    Copy
    z
    
    Note

    The first example is described in the problem statement.

    ac代码为:

    #include<bits/stdc++.h>
    using namespace std;
    char s1[110],s2[110];
    void work(char *s1,int l,int r)
    {
        int temp=r;
        for(int i=l;i<=r;i++) s2[temp--]=s1[i];
        for(int i=l;i<=r;i++) s1[i]=s2[i];
    }
    int main()
    {
        int n;
        scanf("%d",&n);
        scanf("%s",s1+1);
        for(int i=1;i<=n;i++)
        {
            if(n%i==0) work(s1,1,i);    
        }
        printf("%s ",s1+1);
        return 0;
    }

      





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  • 原文地址:https://www.cnblogs.com/csushl/p/9386538.html
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