［水］ZOJ1006

加密方式ciphercode[i] = (plaincode[k*i mod n] - i) mod 28.

n为字符串长度 k为key

题目为给出key和密文

求原来的文字= =

然后=_=因为只是平移就

plain[k*i mod n]= cipher[i]+i 

如果右边笔28大就mod一下就好

密文和原文是一一对应的 （you can assume that untwisting a message always yields a unique result.）

"For those of you with some knowledge of basic number theory or abstract algebra, this will be the case provided that the greatest common divisor of the key k and length n is 1, which it will be for all test cases."

#include <bits/stdc++.h>

using namespace std;
int k;
int a[200],b[200];
int Get(char cc)
{
    if (cc=='_') return 0;
    if (cc=='.') return 27;
    return cc-'a'+1;
}
char Ans(int i)
{
    if (i==0)
        return '_';
    if (i==27)
        return '.';
    return ('a'+i-1);
}
void Gao()
{
    char s[200],s1[200];
    scanf("%s",s);
    int n=strlen(s);
    for (int i=0;i<n;i++)
        a[i]=Get(s[i]);
    for (int i=0;i<n;i++)
    {
        b[(k*i)%n]=a[i]+i;
        if (b[k*i%n]>27)
            b[k*i%n]%=28;
    }
    for (int i=0;i<n;i++)
       cout<<Ans(b[i]);
    cout<<endl;
}
int main()
{
    freopen("a.in","r",stdin);
    while (cin>>k)
    {
        if (k==0)
            break;
        Gao();
    }
    return 0;
}