题解

46 条题解

  • 3
    @ 2018-04-23 23:25:23

    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;
     } 
    
  • 3
    @ 2017-09-05 21:54:15
    #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;
    }
    
  • 2
    @ 2016-12-11 04:15:25

    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;
    }
    ```

    • @ 2017-07-21 09:48:14

      要是我也能这样就好了

  • 2
    @ 2016-11-18 18:23:08
    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.
    
  • 2
    @ 2016-07-11 19:02:38

    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

  • 1
    @ 2021-09-04 14:08:08
    #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;
    }
    
  • 1
    @ 2018-02-06 10:11:15
    #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;
    }
    
  • 1
    @ 2017-10-02 14:32:05
    #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');
            }
        }
    }
    
  • 1
    @ 2016-11-17 15:44:10

    #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;
    }

  • 1
    @ 2016-11-15 21:51:27
    #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;
    }
    
  • 1
    @ 2016-10-14 13:27:33

    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;
    }

  • 0
    @ 2022-08-30 10:01:05
    #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

  • 0
    @ 2020-04-06 15:12:05
    #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;
    }
    
    
  • 0
    @ 2018-02-06 10:08:03
    #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;
    }
    
  • 0
    @ 2016-11-15 21:51:06

    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;
    }

  • 0
    @ 2016-11-15 09:26:46
    #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;
    }
    
  • 0
    @ 2016-11-12 14:23:13
    #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;
    }
    
  • 0
    @ 2016-08-29 16:49:54

    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.

  • 0
    @ 2016-05-15 11:07:46

    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;
    }
    
    • @ 2016-07-11 19:03:06

      不要贴代码,这样子不好

    • @ 2016-10-14 13:27:26

      贴代码只是提供给别人一个参考

  • 0
    @ 2015-11-15 18:39:45

    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

信息

ID
1778
难度
3
分类
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