挺感慨的，一年前读题就读傻了，现在一眼秒了……
基本思路就是先对密钥进行修改，全部变为大写字母，然后根据公式把密文转化为明文。
公式就是c-=(k-'A')（应该还是比较好想的）然后根据原来密文中的大小写形式转化大小写即可。

#include<iostream>
#include<cstring>
using namespace std;
int now;
char k[105],c[1005];
int main()
{
    cin>>k;
    for(int i=0;i<=strlen(k)-1;++i)
        if(k[i]>='a')
            k[i]-=('a'-'A');
    cin>>c;
    for(int i=0;i<=strlen(c)-1;++i)
    {
        if(c[i]>='a')
        {
            c[i]-=(k[now]-'A');
            now++;
            while(c[i]<'a')
                c[i]+=26;
        }
        else
        {
            c[i]-=(k[now]-'A');
            now++;
            while(c[i]<'A')
                c[i]+=26;
        }
        if(now>strlen(k)-1)
            now=0;
    }
    cout<<c;
    return 0;
 } 

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

char key[128];
char info[1024];
char ans[1024];
#define IsUpperCase(ch) ((ch) >= 'A' && (ch) <= 'Z')
#define UpperCase(ch) (IsUpperCase(ch) ? ch : (ch - 'a' + 'A'))

int main() {
    scanf("%s %s", key, info);
    for (int i = 0; i < strlen(info); ++ i)
        ans[i] = ((UpperCase(info[i]) - UpperCase(key[i % strlen(key)])) + 26) % 26 + (IsUpperCase(info[i]) ? 'A' : 'a');
    puts(ans);
    return 0;
}

唉，今非昔比了，当年用了一个小时还只过了7个点，现在分分钟就写好了，只用了20行一次就AC。
```C++
// P1778vigenere密码 https://vijos.org/p/1778
// by Equim 3:50~4:12
#include <cstdio>
#include <cstring>

int main()
{
// INPUT
char key[101];
char ciphertext[1001];
char plaintext[1001];
scanf("%s", key);
scanf("%s", ciphertext);

// PROCESS
strlwr(key);
int keyLength = strlen(key);
int i;
for(i = 0;ciphertext[i]!='\0';i++)
{
plaintext[i] = ciphertext[i] - (key[i%keyLength]-'a');
if ((ciphertext[i] <= 'Z' && plaintext[i] < 'A') || (ciphertext[i] >= 'a' && plaintext[i] < 'a'))
plaintext[i] += 26;
}
plaintext[i] = '\0';

// OUTPUT
printf("%s", plaintext);
return 0;
}
```

program vigenere;

var
    k: string;
    m, c, kc, ans: char;
    i, j, n, l: integer;
    low: boolean;

begin
//    assign(input,'vigenere.in'); assign(output,'vigenere.out');
//    reset(input); rewrite(output);
    readln(k); l:=length(k);
    n := 0;
    while not eoln{(input)} do
    begin
        inc(n);
        read(c);
        low := true; if (ord(c) <= 90) then low := false; c := upcase(c);
        i := n mod l; if (i = 0) then i := l; kc := upcase(k[i]);
        j := ord(c) - ord(kc) + 1; if (j <= 0) then j := j + 26; j := j + 64;
        ans := chr(j); if low then ans := lowercase(ans);
        write(ans);
    end;
//    close(input); close(output);
end.

进行那个公式前 都要转为大写 ,所有的都是大写 最后 并保持字母在明文M中的大小写形式. 枚举一下就ok

#include <bits/stdc++.h>
using namespace std;

int main()
{
    string k,c; cin>>k>>c;
    for (int i=0;i<c.length();i++) {
        int t=(k[i%k.length()]&31)-1;
        c[i]=(c[i]&31)-t>0?c[i]-t:c[i]-t+26;
    }
    cout<<c<<endl;
    return 0;
}

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int k;
int a[110];
char s1[1010], s2[1010];
int change(char c) {
    if(c>='A'&&c<='Z') return c-65;
    if(c>='a'&&c<='z') return c-97;
    return 0;
}
int main() {
    char x;
    x=getchar();
    int j=0;
    while((x>='a'&&x<='z')||(x>='A'&&x<='Z')) {
        j++;
        a[j]=change(x);
        x=getchar();
    }
    scanf("%s", s1);
    strcpy(s2, s1);
    for(int i=0; i<strlen(s1); i++) {
        k=i+1;
        k=(k-1)%j+1;
        for(int jj=1; jj<=a[k]; jj++) {
            s2[i]--;
            if(s2[i]==64) s2[i]+=26;
            if(s2[i]==96) s2[i]+=26;
        }
    }
    printf("%s", s2);
    return 0;
}

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <set>
#include <limits>
#include <string>
#include <sstream>
using namespace std;

const int oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;
const int a_n=26;

int c_size,key_size;
int c_v[1000+1];
int m_v[1000+1];
int key_v[1000+1];
#define key_v(x) key_v[(x-1)%key_size+1]
char c[1000+1];
#define c(x) c[x-1]
char key[1000+1];
#define key(x) key[x-1]

int c_t_i_1(char c)
{
    if ('A'<=c&&c<='Z')
        return c-'A';
    else if ('a'<=c&&c<='z')
        return c-'a';
}

int main()
{
    while (~scanf("%s",key))
    {
        key_size=strlen(key);
        for (int i=1;i<=key_size;i++)
            key_v[i]=c_t_i_1(key(i));
        getchar();
        scanf("%s",c);
        c_size=strlen(c);
        for (int i=1;i<=c_size;i++)
            c_v[i]=c_t_i_1(c(i));
        getchar();
        for (int i=1;i<=c_size;i++)
        {
            m_v[i]=(c_v[i]-key_v(i))%a_n;
            while (m_v[i]<0)
                m_v[i]+=a_n;
            if ('A'<=c(i)&&c(i)<='Z')
                printf("%c",m_v[i]+'A');
            else if ('a'<=c(i)&&c(i)<='z')
                printf("%c",m_v[i]+'a');
        }
    }
}

#include <cstdio>
#include <cstring>

int main(){
//freopen("in.txt","r",stdin);
char key[150],A[1500],B[1500],capital[1500]={0};
scanf("%s%s",key,A);
for(int i=0;i<strlen(key);i++)
if('A'<=key[i]&&key[i]<='Z')
key[i]+=-'A'+'a';
for(int i=0;i<strlen(A);i++)
if('A'<=A[i]&&A[i]<='Z'){
capital[i]=1;
A[i]+=-'A'+'a';
}
for(int i=0;i<strlen(A);i++){
for(char c='a';c<='z';c++)
if((c-'a'+key[i%strlen(key)]-'a')%26==A[i]-'a'){
B[i]=c;
break;
}
}
for(int i=0;i<strlen(A);i++)
if(capital[i])
printf("%c",B[i]-'a'+'A');
else
printf("%c",B[i]);
return 0;
}

#include <iostream>
#include <string>
#include <map>
using namespace std;

string str, strC;
map<char, char> Map[100], _Map[200];

void Init() {
    //大写 
    int nCount = 0;
    char C = 0;
    for (char ch = 'A'; ch <= 'Z'; ch++) {
        nCount = 0;
        C = ch;
        while (nCount < 26) {
            if (C > 'Z') C = 'A';
            Map[ch][nCount + 65] = C;
            
            C ++;
            nCount ++;
        }
    }
    //小写
    for (char ch = 'a'; ch <= 'z'; ch++) {
        nCount = 0;
        C = ch;
        while (nCount < 26) {
            if (C > 'z') C = 'a';
            _Map[ch][nCount + 97] = C;
            
            C ++;
            nCount ++;
        }
    }
}
int main() {
    Init();
    //密匙 密文 
    cin >> str >> strC;
    for (int i = 0; i < strC.length(); i++) {
        char ch_C = strC[i];
        char ch_S = str[i % str.length()];
        //大写 
        if (ch_C >= 'A' && ch_C <= 'Z') {
            if (ch_S < 'A' || ch_S > 'Z') ch_S -= 32;
            for (char ch = 'A'; ch <= 'Z'; ch++) {
                if (Map[ch][ch_S] == ch_C) {
                    cout << ch;
                    break;
                }
            }
        }
        //小写 
        else {
            if (ch_S < 'a' || ch_S > 'z') ch_S += 32;
            for (char ch = 'a'; ch <= 'z'; ch++) {
                if (_Map[ch][ch_S] == ch_C) {
                    cout << ch;
                    break;
                }
            }
        }
    }
    
    return 0;
}

要注意大小写！！！！！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char k[105],m[1005],c[1005];
int main(){
int i,j,l,loth;
cin>>k;
cin>>c;
l=strlen(k);
loth=strlen(c);
for(i=0;i<loth;i++){
int a=i%l,tmp,temp;
if(k[a]>='a'&&k[a]<='z') tmp='a';
else if(k[a]>='A'&&k[a]<='Z') tmp='A';
if(c[i]>='a'&&c[i]<='z') temp='a';
else if(c[i]>='A'&&c[i]<='Z') temp='A';
m[i]=(c[i]-temp-k[a]+tmp+52)%26+temp;
}
cout<<m;
return 0;
}


#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
void turn(char &x);
int main()
{
    char key[1000],secret[1000];
    int len1,len2;
    int i,j;

    cin.getline(key,999,'\n');//输入密钥
    cin.getline(secret,999,'\n');//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
        turn(key[i]);

    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;
}
void turn(char &x)
{
    if((x>='a')&&(x<='z'))
        x-=32;
}

超简单，信息奥赛一本通上有

#include <iostream>         //[2012提高组Day1-A]Vigenère密码
#include <algorithm>
#include <string>
using namespace std;


string Alp = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string alp = "abcdefghijklmnopqrstuvwxyz";
int dik(char k)
{
    if(k >= 'A' && k <= 'Z')
        return k - 'A';
    return k - 'a';
}

char Pass(char c, char k)
{
    if(c >= 'A' && c <= 'Z')
        return Alp[(dik(c) - dik(k) + 26) % 26];
    return alp[(dik(c) - dik(k) + 26) % 26];
}

int main()
{
    string key, C, M;
    cin >> key >> C;
    int Lc = C.length(), Lk = key.length();
    int j = 0;
    for (int i = 0; i < Lc; i++)
    {
        M += Pass(C[i], key[j]);
        j = (j + 1) % Lk;
    }
    cout << M << endl;
    system("pause");
    return 0;
}

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int k;
int a[110];
char s1[1010], s2[1010];
int change(char c) {
    if(c>='A'&&c<='Z') return c-65;
    if(c>='a'&&c<='z') return c-97;
    return 0;
}
int main() {
    char x;
    x=getchar();
    int j=0;
    while((x>='a'&&x<='z')||(x>='A'&&x<='Z')) {
        j++;
        a[j]=change(x);
        x=getchar();
    }
    scanf("%s", s1);
    strcpy(s2, s1);
    for(int i=0; i<strlen(s1); i++) {
        k=i+1;
        k=(k-1)%j+1;
        for(int jj=1; jj<=a[k]; jj++) {
            s2[i]--;
            if(s2[i]==64) s2[i]+=26;
            if(s2[i]==96) s2[i]+=26;
        }
    }
    printf("%s", s2);
    return 0;
}

感觉原题意更好写啊 有一个表的 MAP一下就好了啊

#include <iostream>
#include <string>
#include <map>
using namespace std;

string str, strC;
map<char, char> Map[100], _Map[200];

void Init() {
//大写
int nCount = 0;
char C = 0;
for (char ch = 'A'; ch <= 'Z'; ch++) {
nCount = 0;
C = ch;
while (nCount < 26) {
if (C > 'Z') C = 'A';
Map[ch][nCount + 65] = C;

C ++;
nCount ++;
}
}
//小写
for (char ch = 'a'; ch <= 'z'; ch++) {
nCount = 0;
C = ch;
while (nCount < 26) {
if (C > 'z') C = 'a';
_Map[ch][nCount + 97] = C;

C ++;
nCount ++;
}
}
}
int main() {
Init();
//密匙 密文
cin >> str >> strC;
for (int i = 0; i < strC.length(); i++) {
char ch_C = strC[i];
char ch_S = str[i % str.length()];
//大写
if (ch_C >= 'A' && ch_C <= 'Z') {
if (ch_S < 'A' || ch_S > 'Z') ch_S -= 32;
for (char ch = 'A'; ch <= 'Z'; ch++) {
if (Map[ch][ch_S] == ch_C) {
cout << ch;
break;
}
}
}
//小写
else {
if (ch_S < 'a' || ch_S > 'z') ch_S += 32;
for (char ch = 'a'; ch <= 'z'; ch++) {
if (_Map[ch][ch_S] == ch_C) {
cout << ch;
break;
}
}
}
}

return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//Limits
const int N=100;
const int M=1000;
const int MOD=26;
//Main
char key[N+1];
char cip[M+1];
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    scanf("%s",key+1);
    int keylen=strlen(key+1);
    scanf("%s",cip+1);
    int ciplen=strlen(cip+1);
    int i=1;
    for(int j=1;j<=ciplen;j++)
    {
        int k=tolower(key[i])-'a';
        int c=tolower(cip[j])-'a';
        int m=(c-k+MOD)%MOD+'a';
        if(isupper(cip[j]))m=toupper(m);
        printf("%c",m);
        i++;
        if(i==keylen+1)i=1;
    }
    printf("\n");
    return 0;
}

#include <iostream>
#include <cstring>
using namespace std;
char k[101],a[1001],b[1001],c[1001];
int main()
{
    int size,i,keys;
    gets(k);
    gets(a);
    size=strlen(a);
    keys=strlen(k);
    for(int i=0;i<size;i++)
    {
        if(k[i]>='a'&&k[i]<='z')    k[i]-=32;
        if(a[i]>='a'&&a[i]<='z')    b[i]=a[i]-32;
        else b[i]=a[i];
    }
    for(i=0;i<size;i++)
    {
        int j=i%keys;
        if(b[i]>=k[j])
            c[i]=b[i]+65-k[j];
        else
            c[i]=b[i]+65-k[j]+26;
        if(a[i]>='a'&&a[i]<='z')    c[i]+=32;
        
    }
    cout<<c;
    return 0;
}

先让密钥足够长，然后枚举每一位可能的字母，判断，最后输出即可。
用ansistring
pascal
var k,m,s:ansistring;
i:longint;
c:char;
begin
readln(k);
readln(m);
s:=m;
k:=upcase(k);
m:=upcase(m);
while length(k)<length(m)do k:=k+k;
for i:=1 to length(m)do
for c:='A'to'Z'do
if ord(m[i])=(ord(c)-ord('A')*2+ord(k[i]))mod 26+ord('A')then
begin
m[i]:=c;
break;
end;
for i:=1 to length(m)do
if s[i]=upcase(s[i])then write(m[i])else
write(lowercase(m[i]));
writeln;
end.


网上搜了搜原题是有表的
公式要自己思考出来

#include <cstdio>
#include <cstring>
#include <iostream>

int main(){
    char key[150];
    char classified[1500];
    std::cin>>key>>classified;
    int i=0,len=strlen(key);
    int capital,c; 
    while(classified[i]!='\0'){
        c=key[i%len];
        capital=c>='A'&&c<='Z';
        if(!capital)
            key[i%len]+=-'a'+'A';
            
        c=classified[i];
        capital=c>='A'&&c<='Z';
        if(!capital)
            c=c-'a'+'A';
        for(int x='A';x<='Z';x++)
            if(c==(x-'A'+key[i%len]-'A')%26+'A')
                if(capital)
                    printf("%c",x);
                else
                    printf("%c",x-'A'+'a');
        i++;
    }   
    return 0;
}

评测状态 Accepted
题目 P1778 vigenere密码
递交时间 2015-11-15 18:32:15
代码语言 C++
评测机 VijosEx
消耗时间 45 ms
消耗内存 252 KiB
评测时间 2015-11-15 18:32:17

评测结果

编译成功

foo.cpp: In function 'char qw(char, char)':
foo.cpp:31:1: warning: control reaches end of non-void function [-Wreturn-type]
}
^

测试数据 #0: Accepted, time = 15 ms, mem = 248 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 252 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 248 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 252 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 248 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 252 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 248 KiB, score = 10

测试数据 #7: Accepted, time = 15 ms, mem = 252 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 252 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 252 KiB, score = 10

Accepted, time = 45 ms, mem = 252 KiB, score = 100