题目描述

16 世纪法国外交家 Blaise de Vigenère 设计了一种多表密码加密算法――Vigenère 密

码。Vigenère 密码的加密解密算法简单易用，且破译难度比较高，曾在美国南北战争中为

南军所广泛使用。

在密码学中，我们称需要加密的信息为明文，用 M 表示；称加密后的信息为密文，用

C 表示；而密钥是一种参数，是将明文转换为密文或将密文转换为明文的算法中输入的数据，

记为 k。 在 Vigenère 密码中，密钥 k 是一个字母串，k=k1k2…kn。当明文 M=m1m2…mn时，

得到的密文 C=c1c2…cn，其中 ci=mi®ki，运算®的规则如下表所示：

洛谷 P1079 Vigenère 密码_水题集合

Vigenère 加密在操作时需要注意：

  1. ®运算忽略参与运算的字母的大小写，并保持字母在明文 M 中的大小写形式；

  2. 当明文 M 的长度大于密钥 k 的长度时，将密钥 k 重复使用。

例如，明文 M=Helloworld，密钥 k=abc 时，密文 C=Hfnlpyosnd。

洛谷 P1079 Vigenère 密码_水题集合_02

输入输出格式

输入格式：

 

输入共 2 行。

第一行为一个字符串，表示密钥 k，长度不超过 100，其中仅包含大小写字母。第二行

为一个字符串，表示经加密后的密文，长度不超过 1000，其中仅包含大小写字母。

 

输出格式：

 

输出共 1 行，一个字符串，表示输入密钥和密文所对应的明文。

 

输入输出样例

输入样例#1：
CompleteVictory
Yvqgpxaimmklongnzfwpvxmniytm
输出样例#1：
Wherethereisawillthereisaway

说明

【数据说明】

对于 100%的数据，输入的密钥的长度不超过 100，输入的密文的长度不超过 1000，且

都仅包含英文字母。

NOIP 2012 提高组 第一天 第一题

 

模拟 

#include <cstring>
#include <cstdio>
#define N 1005
#define rep(a,b,c) for(int a=b;a<=c;++a)

int jm[45][45]=
        {{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25},
        {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0},
        {2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1},
        {3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2},
        {4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3},
        {5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4},
        {6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5},
        {7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6},
        {8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7},
        {9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8},
        {10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9},
        {11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10},
        {12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11},
        {13,14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12},
        {14,15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13},
        {15,16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14},
        {16,17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15},
        {17,18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16},
        {18,19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17},
        {19,20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18},
        {20,21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19},
        {21,22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20},
        {22,23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21},
        {23,24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22},
        {24,25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23},
        {25,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}};
char k[N],s[N];
int main()
{
    scanf("%s",k);
    scanf("%s",s);
    int lk=strlen(k),ls=strlen(s),j=0;
    rep(i,0,ls-1)
    {
        int p=k[j]-'A',g=s[i]-'A',f=0;
        if(p>25) p-=32;
        if(g>25) g-=32,f=1;
        rep(k,0,25)
         if(jm[p][k]==g)
          f?putchar(k+'A'+32):putchar(k+'A');
        j=(j+1)%lk;
    }
    return 0;
}

 

我们都在命运之湖上荡舟划桨，波浪起伏着而我们无法逃脱孤航。但是假使我们迷失了方向，波浪将指引我们穿越另一天的曙光。