题解 - Vigenère密码 - 核OJ

幻想一笑而过 LV 9 @ 2023-7-12 9:58:21

#include<bits/stdc++.h>
using namespace std;
string k,C;
char M[130][130]; // 在M[i][i]中，含义为密文i通过密钥j对应的明文
int main(){
    ios::sync_with_stdio(false); // 输入输出加速
    cin.tie(0);
    cout.tie(0);
    cin>>k>>C;
    // a=97 A=65 Z=90 z=122 这是几个重要英文的ascll值
    for(int i=int('A');i<=int('Z');i++){ // i这里是明文
        for(int j=int('A');j<=int('Z');j++){ // j这里是密钥
            int num=i-int('A')+j; // 由题目中的图片找规律计算对应的密文
            while(num>90)num-=26; // 密文的ascll大于Z的话就要回到A
            M[num][j]=M[num][j+32]=char(i); // 存储密文密钥对应的明文
            // 因为忽略转换过程大小写，所以密钥对应的大写值也要存储
        }
    }
    for(int i=int('a');i<=int('z');i++){
        for(int j=int('a');j<=int('z');j++){
            int num=i-int('a')+j;
            while(num>122)num-=26; // 大于z要回到a
            M[num][j]=M[num][j-32]=char(i);
            // 因为忽略转换过程大小写，所以密钥对应的小写值也要存储
        }
    }
    int cnt=0,lenC=C.size(),lenK=k.size();
    for(int i=0;i<lenC;i++){ // 遍历字符串C
        cout<<M[int(C[i])][int(k[cnt++])]; // 输出密文Ci与密钥ki对应的明文Mi
        if(cnt==lenK)cnt=0; // 密钥长度小于密文长度时重新遍历密钥
    }
    return 0;
}

hetao5457389 @ 2024-2-27 6:59:57

w! c!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
陈诗雅 @ 2024-5-28 18:34:36

666

hetao4231912 LV 8 @ 2023-10-2 12:34:10

这一题无非是绕来绕去搞得十分复杂，但思路理清晰后还是比较简单的。

第一种利用设置二维列表。方法复杂度很小，但是你打出这个数组的时间就慢的不得了，这个数组如下

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}};

打完这一个数组，在考试的时候已经花了将近七八分钟。所以在考试时这是没有办法的办法。

但是主函数里的东西明显的就少，注意一下，最好不要经常用万能头文件，这是一个偷懒且不好的习惯，这里需要以下头文件：

#include <cstring>
#include <cstdio>

还需要用到头结点：

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

头结点作用：便于首元结点的处理，最后结点的指针域为空。

以下是主函数：

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;
}

这一串主函数很好理解，不多加赘述。

总体呈现：

#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;
}

第二种方法是大家都比较喜欢的方法：遍历密钥与密文,但是注意密钥、密文的大小写。

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

int main()
{
    char key[1000],secret[1000];
    int len1,len2;
    int i,j;
 
    gets(key);//输入密钥
    gets(secret);//输入密文
    len1=strlen(key);//求密钥长度
    len2=strlen(secret);//求密文长度
 
    for(i=0; i<len1; i++) //将密钥转换为大写字母
       if((key[i]>='a')&&(key[i]<='z'))
        	key[i]-=32;
 
    for(i=0,j=0; i<len2; i++) //从头开始解密
    {
        if(secret[i]<='Z')//大写的情况
        {
            secret[i]=secret[i]-(key[j]-'A');//按规则解密
            j++;
            if(secret[i]<'A')//若明文小于A
                secret[i]='Z'-('A'-secret[i])+1;//从字母表倒序转换
        }
        else//小写的情况
        {
            secret[i]=secret[i]-(key[j]-'A');//按规则解密
            j++;
            if(secret[i]<'a')//若明文小于a
                secret[i]='z'-('a'-secret[i])+1;//从字母表倒序转换
        }
        if(j>len1-1)//若明文长度大于密钥长度，重复使用密钥
            j=0;
    }
    cout<<secret<<endl;
    return 0;
}

这就是两种写法。

AC Code1

#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;
}

AC Code 2

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

int main()
{
    char key[1000],secret[1000];
    int len1,len2;
    int i,j;
 
    gets(key);//输入密钥
    gets(secret);//输入密文
    len1=strlen(key);//求密钥长度
    len2=strlen(secret);//求密文长度
 
    for(i=0; i<len1; i++) //将密钥转换为大写字母
       if((key[i]>='a')&&(key[i]<='z'))
        	key[i]-=32;
 
    for(i=0,j=0; i<len2; i++) //从头开始解密
    {
        if(secret[i]<='Z')//大写的情况
        {
            secret[i]=secret[i]-(key[j]-'A');//按规则解密
            j++;
            if(secret[i]<'A')//若明文小于A
                secret[i]='Z'-('A'-secret[i])+1;//从字母表倒序转换
        }
        else//小写的情况
        {
            secret[i]=secret[i]-(key[j]-'A');//按规则解密
            j++;
            if(secret[i]<'a')//若明文小于a
                secret[i]='z'-('a'-secret[i])+1;//从字母表倒序转换
        }
        if(j>len1-1)//若明文长度大于密钥长度，重复使用密钥
            j=0;
    }
    cout<<secret<<endl;
    return 0;
}

hetao1285761 LV 9 @ 2023-10-28 14:22:38

首先，我们需要定义一个函数，用于执行Vigenère密码的加密解密操作。然后，我们读取输入数据，并调用这个函数来计算明文。最后，我们将结果输出。

#include <iostream>
#include <string>
#include <map>
#include <algorithm>

using namespace std;

// 定义一个map，用于存储密钥和对应的字典序
map<char, int> key_order;

// 初始化密钥和字典序的映射关系
void init_key_order() {
    for (char c = 'a'; c <= 'z'; ++c) {
        key_order[c] = c - 'a';
    }
    for (char c = 'A'; c <= 'Z'; ++c) {
        key_order[c] = c - 'A' + 26;
    }
}

// 执行Vigenère密码的加密解密操作
string vigenère_cipher(const string &m, const string &k) {
    string result;
    int key_len = k.size();
    int plain_len = m.size();
    
    // 初始化密钥和明文的映射关系
    map<int, char> plain_order;
    for (int i = 0; i < plain_len; ++i) {
        char c = m[i];
        int j;
        if (key_order.find(c) != key_order.end()) {
            j = key_order[c] + key_len * i;
        } else {
            j = key_order[tolower(c)] + key_len * i;
        }
        plain_order[j] = c;
    }

    // 将密文转换为明文
    for (int i = 0; i < key_len; ++i) {
        int j = key_order[k[i]];
        if (plain_order.find(j) != plain_order.end()) {
            result.push_back(plain_order[j]);
        }
    }

    return result;
}

int main() {
    init_key_order();
    string key, cipher;
    cin >> key >> cipher;
    string plain = vigenère_cipher(cipher, key);
    cout << plain << endl;
    return 0;
}

这段代码首先定义了一个map，用于存储密钥和对应的字典序。然后定义了一个函数vigenère_cipher，用于执行Vigenère密码的加密解密操作。这个函数首先初始化密钥和明文的映射关系，然后将密文转换为明文。最后，我们读取输入数据，并调用vigenère_cipher函数来计算明文。

1
hetao3563258 LV 10 @ 2024-2-5 9:21:46

发现题目中少了个东西，来喽
0
hetao3413920 LV 3 @ 2024-2-5 13:16:07
1 #include<iostream> #include<cstdio> #include<cstring> using namespace std;

int main() { char key[1000],secret[1000]; int len1,len2; int i,j;
```
gets(key);//输入密钥
gets(secret);//输入密文
len1=strlen(key);//求密钥长度
len2=strlen(secret);//求密文长度

for(i=0; i<len1; i++) //将密钥转换为大写字母
   if((key[i]>='a')&&(key[i]<='z'))
    	key[i]-=32;

for(i=0,j=0; i<len2; i++) //从头开始解密
{
    if(secret[i]<='Z')//大写的情况
    {
        secret[i]=secret[i]-(key[j]-'A');//按规则解密
        j++;
        if(secret[i]<'A')//若明文小于A
            secret[i]='Z'-('A'-secret[i])+1;//从字母表倒序转换
    }
    else//小写的情况
    {
        secret[i]=secret[i]-(key[j]-'A');//按规则解密
        j++;
        if(secret[i]<'a')//若明文小于a
            secret[i]='z'-('a'-secret[i])+1;//从字母表倒序转换
    }
    if(j>len1-1)//若明文长度大于密钥长度，重复使用密钥
        j=0;
}
cout<<secret<<endl;
return 0;
```
} 2 #include<bits/stdc++.h> using namespace std; string k,C; char M[130][130]; // 在M[i][i]中，含义为密文i通过密钥j对应的明文 int main(){ ios::sync_with_stdio(false); // 输入输出加速 cin.tie(0); cout.tie(0); cin>>k>>C; // a=97 A=65 Z=90 z=122 这是几个重要英文的ascll值 for(int i=int('A');i<=int('Z');i++){ // i这里是明文 for(int j=int('A');j<=int('Z');j++){ // j这里是密钥 int num=i-int('A')+j; // 由题目中的图片找规律计算对应的密文 while(num>90)num-=26; // 密文的ascll大于Z的话就要回到A M[num][j]=M[num][j+32]=char(i); // 存储密文密钥对应的明文 // 因为忽略转换过程大小写，所以密钥对应的大写值也要存储 } } for(int i=int('a');i<=int('z');i++){ for(int j=int('a');j<=int('z');j++){ int num=i-int('a')+j; while(num>122)num-=26; // 大于z要回到a M[num][j]=M[num][j-32]=char(i); // 因为忽略转换过程大小写，所以密钥对应的小写值也要存储 } } int cnt=0,lenC=C.size(),lenK=k.size(); for(int i=0;i<lenC;i++){ // 遍历字符串C cout<<M[int(C[i])][int(k[cnt++])]; // 输出密文Ci与密钥ki对应的明文Mi if(cnt==lenK)cnt=0; // 密钥长度小于密文长度时重新遍历密钥 } return 0; } 3 #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; } 4 #include <iostream> #include <string> #include <map> #include <algorithm>

using namespace std;

// 定义一个map，用于存储密钥和对应的字典序 map<char, int> key_order;

// 初始化密钥和字典序的映射关系 void init_key_order() { for (char c = 'a'; c <= 'z'; ++c) { key_order[c] = c - 'a'; } for (char c = 'A'; c <= 'Z'; ++c) { key_order[c] = c - 'A' + 26; } }

// 执行Vigenère密码的加密解密操作 string vigenère_cipher(const string &m, const string &k) { string result; int key_len = k.size(); int plain_len = m.size();
```
// 初始化密钥和明文的映射关系
map<int, char> plain_order;
for (int i = 0; i < plain_len; ++i) {
    char c = m[i];
    int j;
    if (key_order.find(c) != key_order.end()) {
        j = key_order[c] + key_len * i;
    } else {
        j = key_order[tolower(c)] + key_len * i;
    }
    plain_order[j] = c;
}

// 将密文转换为明文
for (int i = 0; i < key_len; ++i) {
    int j = key_order[k[i]];
    if (plain_order.find(j) != plain_order.end()) {
        result.push_back(plain_order[j]);
    }
}

return result;
```
}

int main() { init_key_order(); string key, cipher; cin >> key >> cipher; string plain = vigenère_cipher(cipher, key); cout << plain << endl; return 0; }

ID: 2057
时间: 1000ms
内存: 128MiB
难度: 3
标签: (无)
递交数: 267
已通过: 135
上传者: TomAnderson

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Vigenère密码

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5 条题解

Vigenère密码

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