题解

130 条题解

  • 0
    @ 2021-10-26 16:59:17

    数据有亿点水,只需要判断首字母

    #include<iostream>
    #include<cstdio>
    using namespace std;
    int n;
    string a1,a2,a3;
    int jud(string a,string b,string c)
    {
        int i,j;
        for(i=0,j=n-1;i<=n-1,j>=0;i++,j--)
        {
            int t1=a[i]-'a'+1;
            int t2=b[i]-'a'+1;
            int t3=c[j]-'a'+1;
            int mark=0;
            for(int k=0;k<=26;k++)
            {
                if(t1==(t3+k-1)%26+1 && t2==(t3-k-1+26)%26+1)//玄妙的取模运算,可以省掉四个判断
                {
                mark=1;break;}
                if(t2==(t3+k-1)%26+1 && t1==(t3-k-1+26)%26+1)
                {
                mark=1;break;}
            }
            if(mark)
            {
            continue;
            }
            else
            return 0;
        }
        return 1;
    }
    void output(string a)//反向输出找到的倒序字串
    {
        int i;
        for(i=n-1;i>=0;i--)
        cout<<a[i];
    }
    int main()
    {
        cin>>n;
        cin>>a1>>a2>>a3;
        if(jud(a1,a2,a3))//分别判断倒序的字串究竟是哪个
        {
            output(a3);
            return 0;
        }
        if(jud(a2,a3,a1))
        {
            output(a1);
            return 0;
        }
        if(jud(a3,a1,a2))
        {
            output(a2);
            return 0;
        }
        //system("pause");
    }
    

    芜湖,废了我1个小时

  • 0
    @ 2021-05-20 15:47:50

    ``java

    import java.io.BufferedReader;
    import java.io.InputStreamReader;
    import java.util.Scanner;

    /**
    * @author wpx
    * @version V1.0
    * @Package com.algorithm
    * @date 2021/5/20 9:40
    */
    public class Main {
    public static void main(String[] args) {
    Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    sc.nextLine();
    String strOne = sc.nextLine().trim();
    String strTwo = sc.nextLine().trim();
    String strThree = sc.nextLine().trim();
    // a = 97 z = 122
    // 先找出前后移动字母的加密方式
    // 先比较 第一个和第二个字符
    String realStr = null;
    realStr = hasEqualGap(strOne, strTwo);
    if(realStr != null && realStr.equals(getTrueValue(strThree))){
    System.out.println(realStr);
    return;
    }
    realStr = hasEqualGap(strOne, strThree);
    if(realStr != null && realStr.equals(getTrueValue(strTwo))){
    System.out.println(realStr);
    return;
    }

    realStr = hasEqualGap(strTwo, strThree);
    if(realStr != null && realStr.equals(getTrueValue(strOne))){
    System.out.println(realStr);
    return;
    }
    }
    private static String hasEqualGap(String strOne, String strTwo){
    Integer gap = null;
    boolean hasEqualGap = true;
    StringBuilder builder = new StringBuilder();
    for(int i = 0; i < strOne.length(); i++){
    final char charOne = strOne.charAt(i);
    final char charTwo = strTwo.charAt(i);
    int nowGap = Math.abs(charOne - charTwo) ;
    if(nowGap > 12){
    nowGap = Math.abs(26 - nowGap);
    }
    if(gap == null) {
    gap = nowGap;
    } else if(gap != nowGap){
    return null;
    }
    if(gap == nowGap){
    // 假设 strOne 为向前移动
    if(getTrueChar((char)(charOne - gap/2)) == getTrueChar((char)(charTwo + gap/2)) ) {
    builder.append(getTrueChar((char)(charOne - gap/2)));
    } else if (getTrueChar((char)(charOne + gap/2)) == getTrueChar((char)(charTwo - gap/2))) {
    builder.append(getTrueChar((char)(charOne + gap/2)));
    } else {
    return null;
    }

    }
    }
    return builder.toString();
    }

    public static char getTrueChar(char charValue){
    if(charValue > 122) {
    return (char) (96 + (charValue - 122));
    } else if(charValue < 97) {
    return (char) (123 + (charValue - 97));
    } else {
    return charValue;
    }
    }

    private static String getTrueValue(String str) {
    StringBuilder stringBuilder = new StringBuilder(str);
    stringBuilder.reverse();
    return stringBuilder.toString();
    }
    }

  • 0
    @ 2020-11-05 11:33:55

    #include <stdio.h>
    #include <string.h>
    #include <math.h>

    //翻转字符串函数
    void versech(char a[],int N){
    int i;
    char ch;
    for(i=0;i<N/2;i++){
    ch=a[i];
    a[i]=a[N-1-i];
    a[N-1-i]=ch;
    }
    }

    //字符串相加函数
    void add_str(char a[],char b[],char c[],int N){
    int i;
    for(i=0;i<N;i++){
    //若某两个字符之差绝对值在12以内,说明没有分布在‘a’的两侧
    if(abs((a[i]-'a')%26-(b[i]-'a')%26)<=12)
    c[i]=(((a[i]-'a')%26+(b[i]-'a')%26)/2)%26+'a';
    //否则在'a'的两侧需将两字符之和加上26
    else
    c[i]=(((a[i]-'a')%26+(b[i]-'a')%26+26)/2)%26+'a';
    }
    //注意,若使用字符串函数,要记得字符串最后加'\0',否则使用字符串函数会出错
    c[N]='\0';
    }

    int main()
    {
    char a1[10001];
    char a2[10001];
    char a3[10001];
    char add[10001];
    int n;
    scanf("%d",&n);
    scanf("%s",a1);
    scanf("%s",a2);
    scanf("%s",a3);
    //翻转a1
    versech(a1,n);
    add_str(a2,a3,add,n);
    if(!strcmp(a1,add))
    printf("%s\n",a1);
    else{
    //否则将a1翻转回来,继续翻转a2
    versech(a1,n);
    versech(a2,n);
    add_str(a1,a3,add,n);
    if(!strcmp(a2,add))
    printf("%s\n",a2);
    else{
    //否则将a2翻转回来,继续翻转a3
    versech(a2,n);
    versech(a3,n);
    add_str(a1,a2,add,n);
    if(!strcmp(add,a3))
    printf("%s\n",a3);
    }
    }
    return 0;

    }

  • 0
    @ 2020-03-29 12:38:27

    #include<iostream>
    #include<cstdlib>
    #include<string>
    #include<array>
    #include<vector>
    using namespace std;
    string decrypt1(string a, int k,int n)
    {
    int temp;
    const array<char, 27>letters = { '0','a','b','c','d','e','f','g','h','i','j','k','l','m','n',
    'o','p','q','r','s','t','u','v','w','x' ,'y','z' };
    for (int i = 0; i < n; i++)
    {
    for (int s = 0; s < 27; s++)
    {
    if (a[i] == letters[s])
    {
    temp = s;
    break;
    }
    }
    if (temp + k <= 26)
    a[i] = letters[temp + k];
    else
    a[i] = letters[temp + k - 26];
    }
    return a;
    }
    string decrypt2(string a, int k, int n)
    {
    int temp;
    const array<char, 27>letters = { '0','a','b','c','d','e','f','g','h','i','j','k','l','m','n',
    'o','p','q','r','s','t','u','v','w','x' ,'y','z' };
    for (int i = 0; i < n; i++)
    {
    for (int s = 0; s < 27; s++)
    {
    if (a[i] == letters[s])
    {
    temp = s;
    break;
    }
    }
    if (temp - k > 0)
    a[i] = letters[temp - k];
    else
    a[i] = letters[temp - k + 26];
    }
    return a;
    }
    string decrypt3(string a, int n)
    {
    vector<char>temp(n);
    for (int k = 0; k < n; k++)
    {
    temp[n - k - 1] = a[k];
    }
    for (int i = 0; i < n; i++)
    {
    a[i] = temp[i];
    }
    return a;
    }
    int main()
    {
    int n;
    string s1, s2, s3, s4;
    cin >> n >> s1 >> s2 >> s3;
    for (int j = 0; j <= 6; j++)
    {
    if (decrypt1(s1, j, n) == decrypt2(s2, j, n)&& decrypt1(s1, j, n)==decrypt3(s3,n))
    {
    s4 = decrypt1(s1, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s2, j, n) == decrypt2(s3, j, n) && decrypt1(s2, j, n) == decrypt3(s1, n))
    {
    s4 = decrypt1(s2, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s3, j, n) == decrypt2(s1, j, n) && decrypt1(s3, j, n) == decrypt3(s2, n))
    {
    s4 = decrypt1(s3, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s2, j, n) == decrypt2(s1, j, n) && decrypt1(s2, j, n) == decrypt3(s3, n))
    {
    s4 = decrypt1(s2, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s3, j, n) == decrypt2(s2, j, n) && decrypt1(s3, j, n) == decrypt3(s1, n))
    {
    s4 = decrypt1(s3, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s1, j, n) == decrypt2(s3, j, n) && decrypt1(s1, j, n) == decrypt3(s2, n))
    {
    s4 = decrypt1(s1, j, n);
    cout << s4;
    break;
    }
    }
    return 0;
    }

  • 0
    @ 2020-03-29 12:37:55

    #include<iostream>
    #include<cstdlib>
    #include<string>
    #include<array>
    #include<vector>
    using namespace std;
    string decrypt1(string a, int k,int n)
    {
    int temp;
    const array<char, 27>letters = { '0','a','b','c','d','e','f','g','h','i','j','k','l','m','n',
    'o','p','q','r','s','t','u','v','w','x' ,'y','z' };
    for (int i = 0; i < n; i++)
    {
    for (int s = 0; s < 27; s++)
    {
    if (a[i] == letters[s])
    {
    temp = s;
    break;
    }
    }
    if (temp + k <= 26)
    a[i] = letters[temp + k];
    else
    a[i] = letters[temp + k - 26];
    }
    return a;
    }
    string decrypt2(string a, int k, int n)
    {
    int temp;
    const array<char, 27>letters = { '0','a','b','c','d','e','f','g','h','i','j','k','l','m','n',
    'o','p','q','r','s','t','u','v','w','x' ,'y','z' };
    for (int i = 0; i < n; i++)
    {
    for (int s = 0; s < 27; s++)
    {
    if (a[i] == letters[s])
    {
    temp = s;
    break;
    }
    }
    if (temp - k > 0)
    a[i] = letters[temp - k];
    else
    a[i] = letters[temp - k + 26];
    }
    return a;
    }
    string decrypt3(string a, int n)
    {
    vector<char>temp(n);
    for (int k = 0; k < n; k++)
    {
    temp[n - k - 1] = a[k];
    }
    for (int i = 0; i < n; i++)
    {
    a[i] = temp[i];
    }
    return a;
    }
    int main()
    {
    int n;
    string s1, s2, s3, s4;
    cin >> n >> s1 >> s2 >> s3;
    for (int j = 0; j <= 6; j++)
    {
    if (decrypt1(s1, j, n) == decrypt2(s2, j, n)&& decrypt1(s1, j, n)==decrypt3(s3,n))
    {
    s4 = decrypt1(s1, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s2, j, n) == decrypt2(s3, j, n) && decrypt1(s2, j, n) == decrypt3(s1, n))
    {
    s4 = decrypt1(s2, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s3, j, n) == decrypt2(s1, j, n) && decrypt1(s3, j, n) == decrypt3(s2, n))
    {
    s4 = decrypt1(s3, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s2, j, n) == decrypt2(s1, j, n) && decrypt1(s2, j, n) == decrypt3(s3, n))
    {
    s4 = decrypt1(s2, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s3, j, n) == decrypt2(s2, j, n) && decrypt1(s3, j, n) == decrypt3(s1, n))
    {
    s4 = decrypt1(s3, j, n);
    cout << s4;
    break;
    }
    if (decrypt1(s1, j, n) == decrypt2(s3, j, n) && decrypt1(s1, j, n) == decrypt3(s2, n))
    {
    s4 = decrypt1(s1, j, n);
    cout << s4;
    break;
    }
    }
    return 0;
    }

  • 0
    @ 2019-09-22 21:04:56

    #include<iostream>
    #include<string>
    #include<algorithm>
    using namespace std;
    int n;
    string h1(string s)//方法1
    { int i,t;
    string f;
    for(i=0;i<n/2;i++)
    { f[0]=s[i];
    s[i]=s[n-i-1];
    s[n-i-1]=f[0];
    }
    return s;
    }
    string h2(string s,int k)//方法2 逆向
    { int i,t;
    for(i=0;i<n;i++)
    for(t=0;t<k;t++)
    if(s[i]-1<'a') s[i]='z';
    else s[i]=s[i]-1;
    return s;
    }
    string h3(string s,int k)//方法3 逆向
    { int i,t;
    for(i=0;i<n;i++)
    for(t=0;t<k;t++)
    if(s[i]+1>'z') s[i]='a';
    else s[i]=s[i]+1;
    return s;
    }
    int main()
    { int i;
    string s1,s2,s3;
    cin>>n;
    cin>>s1>>s2>>s3;
    for(i=0;i<=6;i++)//方法全排列
    { if(h1(s1)==h2(s2,i)&&h2(s2,i)==h3(s3,i))
    { cout<<h1(s1);
    break;
    }
    if(h1(s1)==h2(s3,i)&&h2(s3,i)==h3(s2,i))
    { cout<<h1(s1);
    break;
    }
    if(h1(s2)==h2(s1,i)&&h2(s1,i)==h3(s3,i))
    { cout<<h1(s2);
    break;
    }
    if(h1(s2)==h2(s3,i)&&h2(s3,i)==h3(s1,i))
    { cout<<h1(s2);
    break;
    }
    if(h1(s3)==h2(s1,i)&&h2(s1,i)==h3(s2,i))
    { cout<<h1(s3);
    break;
    }
    if(h1(s3)==h2(s2,i)&&h2(s2,i)==h3(s1,i))
    { cout<<h1(s3);
    break;
    }
    }

    }

  • 0
    @ 2019-06-16 15:21:37

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

    string dao(string x,int n){
    string ans(n, '0');
    int i,j;
    for( i=0,j=x.length()-1;i<x.length();i++,j--){
    ans[i]=x[j];
    }
    return ans;
    }

    int main(){
    int n;
    cin>>n;
    string a,b,c;
    cin>>a>>b>>c;
    if((a[0]-dao(b,n)[0]+c[0]-dao(b,n)[0])%26==0){
    cout<<dao(b,n);
    }else if((a[0]-dao(c,n)[0]+b[0]-dao(c,n)[0])%26==0){
    cout<<dao(c,n);
    }else if((b[0]-dao(a,n)[0]+c[0]-dao(a,n)[0])%26==0){
    cout<<dao(a,n);
    }
    }

  • 0
    @ 2018-09-07 18:29:55

    #include<iostream>
    #include<string>
    using namespace std;
    string all("abcdefghijklmnopqrstuvwxyz");
    string zb(int p,string q,int n) {
    for(int i=0; i<n; i++) {
    int x=all.find(q[i]);
    x=(x+p+26)%26;
    q[i]=all[x];
    }
    return q;
    }
    string dd(string o,int n) {
    for(int i=1; i<=n/2; i++) {
    char t=o[i-1];
    o[i-1]=o[n-i];
    o[n-i]=t;
    }
    return o;
    }
    int main() {
    string a,b,c,temp;
    int n,k;
    cin>>n;
    getline(cin,temp);
    getline(cin,a);
    getline(cin,b);
    getline(cin,c);
    for(k=0; k<=6; k++) {
    if((zb(k,dd(a,n),n)==b&&zb(-k,dd(a,n),n)==c)||(zb(k,dd(a,n),n)==c&&zb(-k,dd(a,n),n)==b)){cout<<dd(a,n);k=k+7;}
    else if((zb(k,dd(b,n),n)==a&&zb(-k,dd(b,n),n)==c)||(zb(k,dd(b,n),n)==c&&zb(-k,dd(b,n),n)==a)){cout<<dd(b,n);k=k+7;}
    else if((zb(k,dd(c,n),n)==b&&zb(-k,dd(c,n),n)==a)||(zb(k,dd(c,n),n)==a&&zb(-k,dd(c,n),n)==b)){cout<<dd(c,n);k=k+7;}
    }
    return 0;
    }

  • 0
    @ 2018-05-25 15:11:01
    #include<iostream>
    #include<cstdio>
    using namespace std;
    const int MAXN=10000+50;
    char str[4][MAXN];
    int d1[4]= {0,1,1,2};
    int d2[4]= {0,2,3,3};
    int d3[4]= {0,3,2,1};
    int n;
    bool ok(char a[],char b[],int f) {
        for(int i=0; i<n; i++) {
            char x;
            if(a[i]+f<'a')x='z'+f+a[i]-'a'+1;
            else if(a[i]+f>'z')x=a[i]+f-'z'+'a'-1;
            else x=a[i]+f;
            if(x!=b[n-i-1])return false;
        }
        return true;
    }
    int main() {
        scanf("%d",&n);
        scanf("%s%s%s",str[1],str[2],str[3]);
        for(int i=1; i<=3; i++) {
            for(int k=0; k<=6; k++) {
                if(ok(str[d1[i]],str[d3[i]],-k)&&ok(str[d2[i]],str[d3[i]],k)) {
                    for(int j=n-1; j>=0; j--) {
                        printf("%c",str[d3[i]][j]);
                    }
                    return 0;
                }
                if(ok(str[d1[i]],str[d3[i]],k)&&ok(str[d2[i]],str[d3[i]],-k)) {
                    for(int j=n-1; j>=0; j--) {
                        printf("%c",str[d3[i]][j]);
                    }
                    return 0;
                }
            }
        }
        return 0;
    }
    
  • 0
    @ 2018-02-20 16:23:37

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

    using namespace std;

    bool decideStr(string a,string b){
    if(a == b){
    return true;
    }
    else return false;
    }

    char frontChar(char c){
    if(c == 'a'){
    return 'z';
    }
    else{
    return c - 1;
    }
    }

    char backChar(char c){
    if(c == 'z'){
    return 'a';
    }
    else{
    return c + 1;
    }
    }

    string getStr(string a,string b){
    string ans;
    int len = a.size();
    int leng = 0;
    for(int i = 0; i < len ; i++){
    leng = ans.size();
    char f = a[i];
    char n = b[i];
    int tryT = 0;
    while(tryT < 7){
    if(f == n){
    ans = ans + f;
    break;
    }
    else{
    f = frontChar(f);
    n = backChar(n);
    tryT++;
    }
    }
    if(leng != ans.size()){
    continue;
    }
    tryT = 0;
    f = a[i];
    n = b[i];
    while(tryT < 7){
    if(f == n){
    ans = ans + f;
    break;
    }
    else{
    f = backChar(f);
    n = frontChar(n);
    tryT++;
    }
    }
    }
    return ans;
    }

    string turnStr(string a){
    string ans;
    deque<char> que;
    int len = a.size();
    for(int i = 0;i < len;i ++){
    char temp = a[i];
    que.push_back(temp);
    }
    while(!que.empty()){
    ans = ans + que.back();
    que.pop_back();
    }
    return ans;
    }

    int main(){
    int meishayong;
    cin >> meishayong;
    string one,two,three;
    cin >> one >> two >> three;
    if(decideStr(turnStr(one),getStr(two,three))){
    cout << turnStr(one) << endl;
    }
    else if(decideStr(turnStr(two),getStr(one,three))){
    cout << turnStr(two) << endl;
    }
    else if(decideStr(turnStr(three),getStr(one,two))){
    cout << turnStr(three) << endl;
    }
    return 0;
    }

  • 0
    @ 2017-11-22 17:32:09
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <functional>
    
    using namespace std;
    using namespace placeholders;
    
    const int maxN = 1e4 + 5;
    
    char encStr[3][maxN];
    char transStr[3][maxN];
    int len;
    
    void reverseStr(int srcIdx)
    {
        reverse_copy(encStr[srcIdx], encStr[srcIdx] + len, transStr[0]);
    }
    
    void forwardStr(int srcIdx, int destIdx, int step)
    {
        transform(encStr[srcIdx], encStr[srcIdx] + len, transStr[destIdx], [&] (char ch) -> char {
            return (ch - 'a' + step + 26) % 26 + 'a';
        });
    }
    
    void input()
    {
        scanf("%d", &len);
        for (int i = 0; i < 3; i++)
            scanf("%s", encStr[i]);
    }
    
    void solve()
    {
        int idx[] = { 0, 1, 2 };
    
        for (int i = 0; i < 6; i++)
        {
            for (int step = 0; step <= 6; step++)
            {
                function<void(void)> funcs[] = {
                    bind(reverseStr, idx[0]),
                    bind(forwardStr, idx[1], 1, step),
                    bind(forwardStr, idx[2], 2, -step)
                };
                for (auto& fn: funcs)
                    fn();
                if (strcmp(transStr[0], transStr[1]) == 0 && strcmp(transStr[1], transStr[2]) == 0)
                {
                    printf("%s", transStr[0]);
                    return;
                }
            }
            next_permutation(idx, idx + 3);
        }
    }
    
    int main()
    {
        input();
        solve();
        return 0;
    }
    
    

    暴力枚举即可(顺便测试一下C++11的新特性)

  • 0
    @ 2017-10-24 20:16:13
    n=int(raw_input())
    s="zabcdefghijklmnopqrstuvwxyz"
    s1=raw_input()
    s2=raw_input()
    s3=raw_input()
    def sol(ss1,ss2,ss3):
        ans_temp=""
        ans_temp2=""
        ans_temp3=""
        for i in range(n):
            ans_temp=ans_temp+ss1[n-i-1]
        for k in range(7):
            ans_temp2=""
            ans_temp3=""
            for i in range(n):
                ans_temp2=ans_temp2+s[((ord(ss2[i])-96-k+26)%26)]
                ans_temp3=ans_temp3+s[((ord(ss3[i])-96+k+26)%26)]
                if (ans_temp==ans_temp2) and (ans_temp==ans_temp3):
                    print ans_temp
                    exit()
    sol(s1,s2,s3)
    sol(s1,s3,s2)
    sol(s2,s1,s3)
    sol(s2,s3,s1)
    sol(s3,s1,s2)
    sol(s3,s2,s1)
    
  • 0
    @ 2017-09-25 15:40:45

    代码很长,比较容易。

    #include<iostream>
    
    using namespace std;
    
    int n;
    string a, b, c;
    
    string change(string s, int offset, int to);
    string swap2(string s);
    bool check(string a, string b, string c);
    
    int main() {
        cin >> n;
        cin >> a;
        cin >> b;
        cin >> c;
        
        if (check(a, b, c)) {
        } else if (check(a, c, b)) {    
        } else if (check(b, a, c)) {
        } else if (check(b, c, a)) {
        } else if (check(c, a, b)) {
        } else if (check(c, b, a)) {
      }
        
        return 0;
    }
    
    bool check(string a, string b, string c) {
        string a1, b1, c1;
        
        a1 = swap2(a);
        
        for (int k = 0; k <= 6; k++) {
            b1 = change(b, k, -1);
            c1 = change(c, k, 1);
            
            if (a1 == b1 && b1 == c1 && a1 == c1) {
                cout << a1 << "\n";
                return true;
            }
        }
        
        return false;
    }
    
    string swap2(string s) {
        string t = s;
        
        for (int i = 0; i < n; i++)
            t[i] = s[n - 1 - i];
        
        return t;
    }
    
    string change(string s, int offset, int to) {
        string t = s;
        
        if (offset != 0) {
            for (int i = 0; i < n; i++) {
          int j = offset;
          
          while (j) {
            t[i] = t[i] + to;
    
            if (t[i] > 'z') {
              t[i] = 'a';
            }
            if (t[i] < 'a') {
              t[i] = 'z';
            }
    
            j--;
          }
            }
      }
        
        return t;
    }
    
  • 0
    @ 2017-05-17 20:21:33

    int main()
    {
    int N;
    string String[3];
    cin >> N;
    for (int i = 0; i < 3; i++){
    cin >> String[i] ;
    }

    for (int i = 0; i < 3; i++){
    int u = (String[(i + 1) % 3][0] + String[(i + 2) % 3][0] - String[i][N - 1 - 0] - String[i][N - 1 - 0]) % 26;
    if ((u > 6) || (u < -6)){
    continue;
    }
    for (int ii = 1; ii < N; ii++){
    if (u != (String[(i + 1) % 3][ii] + String[(i + 2) % 3][ii] - String[i][N - 1 - ii] - String[i][N - 1 - ii]) % 26){
    u = -9999;
    break;
    }
    }
    if (-9999 != u){
    reverse(String[i].begin(), String[i].end());
    cout << String[i];
    break;
    }
    }
    return 0;
    }

  • 0
    @ 2017-05-07 14:18:34

    #include<iostream>
    using namespace std;

    int judge(char* s1,char* s2,char* s3,int n)
    {
    char *s_new = new char[n];
    for (int i = 0;i<=n/2;i++)
    {
    s_new[i] = s1[n-i-1];
    s_new[n-i-1] = s1[i];
    }
    int flag = 0;
    for(int i = 0;i<n;i++)
    {
    if(s2[i]-s3[i]>12 && flag != -1)
    {
    flag = 1;
    if('z'-s2[i]<=s3[i]-'a'){
    if(s_new[i]!=(s2[i]+s3[i]-26)/2)
    return 0;
    }else{
    if(s_new[i]!=(s2[i]+s3[i]+26)/2)
    return 0;
    }
    }else if(s2[i]-s3[i]<-12 && flag != 1){
    flag = -1;
    if('z'-s3[i]<=s2[i]-'a'){
    if(s_new[i]!=(s2[i]+s3[i]-26)/2)
    return 0;
    }else{
    if(s_new[i]!=(s2[i]+s3[i]+26)/2)
    return 0;
    }
    }else if(s2[i]-s3[i]<=12 && s2[i]-s3[i]>=-12){
    if(s_new[i]!=(s2[i]+s3[i])/2)
    return 0;
    }else{
    return 0;
    }
    }
    return 1;
    }

    void reverse(char* s,int n)
    {
    for(int i = 0;i<n;i++)
    {
    cout<<s[n-i-1];
    }
    return;
    }

    int main()
    {
    //Read
    int n;
    cin>>n;
    char* s1 = new char[n];
    char* s2 = new char[n];
    char* s3 = new char[n];
    cin>>s1>>s2>>s3;

    //Count
    if(judge(s1,s2,s3,n))
    {
    reverse(s1,n);
    }else if(judge(s2,s1,s3,n)){
    reverse(s2,n);
    }else if(judge(s3,s1,s2,n)){
    reverse(s3,n);
    }else{
    cout<<"No anwser"<<endl;
    }
    return 0;
    }

  • 0
    @ 2017-05-06 17:45:35

    #include <iostream>
    #include <cstdio>
    #include <string.h>
    #include <cmath>
    using namespace std;
    const int maxn = 10010;
    char a[maxn];
    char b[maxn];
    char c[maxn];
    bool fun(int t,int a,int b){
    if(((a+b)-2*t)%26==0)
    return false;
    return true;
    }
    int main(){
    int n;
    scanf("%d",&n);
    scanf("%s%s%s",a,b,c);
    int t1,t2,t3,t,a1,a2,a3;
    t1 = t2 = t3 = 1;
    for(int i=0;i<n;++i){
    a1 = a[i] - 'a';
    a2 = b[i] - 'a';
    a3 = c[i] - 'a';
    if(t1){
    t = a[n-i-1] - 'a';
    // cout<<t<<" "<<a2<<" "<<a3<<endl;
    if(fun(t,a2,a3)){
    t1 = 0;
    }
    // cout<<t1<<endl;
    }
    if(t2){
    t = b[n-i-1] - 'a';
    // cout<<t<<" "<<a1<<" "<<a3<<endl;
    if(fun(t,a1,a3)){
    t2 = 0;
    }
    // cout<<t2<<endl;
    }
    if(t3){
    t = c[n-i-1] - 'a';
    // cout<<t<<" "<<a2<<" "<<a1<<endl;
    if(fun(t,a2,a1)){
    t3 = 0;
    }
    // cout<<t3<<endl;
    }
    if(t1+t2+t3==1)
    break;
    }
    if(t1==1){
    for(int i=n-1;i>=0;--i)
    cout<<a[i];
    }else if(t2==1){
    for(int i=n-1;i>=0;--i)
    cout<<b[i];
    }else{
    for(int i=n-1;i>=0;--i)
    cout<<c[i];
    }
    cout<<endl;

    return 0;
    }

  • 0
    @ 2017-02-19 21:48:04

    #include<iostream>
    #include<algorithm>
    #include<string>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<set>
    #include<stack>
    #include<sstream>
    #include<map>
    #include<vector>
    #include<queue>
    using namespace std;
    const int N=10010;
    char a1[N],a2[N],a3[N];
    int length;
    void reverse(char *c,int n){
    int i=0,j=n-1;
    while(i<j)
    {
    swap(c[i],c[j]);
    i++;j--;
    }
    }

    void change(char *c,int n,int k){
    for(int i=0;i<n;i++)
    {
    if(c[i]+k>'z') c[i]=c[i]+k-26;
    else if(c[i]+k<'a') c[i]=c[i]+k+26;
    else c[i]+=k;
    }
    }
    void sl(char *a,char *b,char *c)
    {
    reverse(a,length);
    for(int k=0;k<=6;k++)
    {
    change(b,length,k);
    change(c,length,-k);
    if(!strcmp(b,c)&&!strcmp(a,c))
    {cout<<a<<endl;exit(0);}
    change(b,length,-2*k);
    change(c,length,2*k);
    if(!strcmp(b,c)&&!strcmp(a,c))
    {cout<<a<<endl;exit(0);}
    change(b,length,k);
    change(c,length,-k);
    }
    reverse(a,length);
    }
    int main(){
    cin>>length;
    scanf("%s%s%s",a1,a2,a3);
    sl(a1,a2,a3);
    sl(a2,a3,a1);
    sl(a3,a2,a1);
    return 0;
    }

  • 0
    @ 2017-01-24 09:23:01

    var
    a,b,c,s:ansistring;
    n,i:longint;
    begin
    readln(n);readln(a);readln(b);readln(c);
    if (ord(a[1])+ord(b[1])-2*ord(c[n]))mod 26=0 then s:=c
    else if (ord(a[1])+ord(c[1])-2*ord(b[n]))mod 26=0 then s:=b
    else s:=a;
    for i:=n downto 1 do write(s[i]);
    end.

  • 0
    @ 2017-01-19 11:27:00

    #include<iostream>
    #include<string>
    #include<cstdio>
    using namespace std;
    string sw(string s2)
    {
    string s1="";
    for(int i=0;i<s2.size();i++)
    {
    s1=s2[i]+s1;
    }
    return s1;
    }
    bool bd(string s3,string s4,int k)
    {
    for(int i=0;i<s4.size();i++)
    {
    if(s4[i]-k>='a'){if ((s4[i]-k)!=s3[i])return 0;}
    else
    {if((s4[i]-k+26)!=s3[i])return 0;}
    }
    return 1;
    }

    bool bg(string s3,string s4,int k)
    {
    for(int i=0;i<s4.size();i++)
    {
    if(s4[i]+k<='z'){if ((s4[i]+k)!=s3[i])return 0;}
    else
    {if((s4[i]+k-26)!=s3[i])return 0;}
    }
    return 1;
    }

    int main()
    {
    string sf,sc,st,s1,s2,s3;
    int n;
    scanf("%d",&n);
    cin>>sf>>sc>>st;
    s1=sw(sf);
    s2=sw(sc);
    s3=sw(st);
    for(int k=0;k<=6;k++)
    {
    if((bd(s1,sc,k)&&bg(s1,st,k))||(bd(s1,st,k)&&bg(s1,sc,k))){cout<<s1<<"\n";break;}
    if((bd(s2,sf,k)&&bg(s2,st,k))||(bd(s2,st,k)&&bg(s2,sf,k))){cout<<s2<<"\n";break;}
    if((bd(s3,sf,k)&&bg(s3,sc,k))||(bd(s3,sc,k)&&bg(s3,sf,k))){cout<<s3<<"\n";break;}
    }
    return 0;
    }

  • 0
    @ 2016-10-26 18:59:59
    #include<iostream>
    #include<algorithm>
    using namespace std;
    char s1[5][12000],s2[5][12000];
    int n;
    void case1(int x){
        for(int j=1;j<=n;j++)
            s2[x][j]=s1[x][j];
            for(int i=1;i<=n/2;i++){
                swap(s2[x][i],s2[x][n-i+1]);
            }
    }
    void case2(int x,int k){
        for(int j=1;j<=n;j++)
            s2[x][j]=s1[x][j];
        for(int i=1;i<=n;i++){
            s2[x][i]+=k;
            if(s2[x][i]>'z')s2[x][i]-=26;
        }
    }
    void case3(int x,int k){
        for(int j=1;j<=n;j++)
            s2[x][j]=s1[x][j];
        for(int i=1;i<=n;i++){
            s2[x][i]-=k;
            if(s2[x][i]<'a')s2[x][i]+=26;
        }
    }
    bool ok(){
        for(int i=1;i<=n;i++)
            if(s2[1][i]!=s2[2][i]||s2[2][i]!=s2[3][i])return false;
        return true;
    }
    void solve(){
        for(int k=0;k<7;k++)
            for(int i=0;i<3;i++){
                case1(i+1);
                case2((i+1)%3+1,k);
                case3((i+2)%3+1,k);
                if(ok())return;
                case3((i+1)%3+1,k);
                case2((i+2)%3+1,k);
                if(ok())return;
            }
    }
    int main(){
        ios::sync_with_stdio(false);
        cin>>n;
        for(int i=1;i<4;i++)
            for(int j=1;j<=n;j++)
                cin>>s1[i][j];
        solve();
        for(int i=1;i<=n;i++)cout<<s2[1][i];
        cout<<endl;
        return 0;
    }
    

信息

ID
1449
难度
6
分类
字符串 | 模拟 点击显示
标签
递交数
6903
已通过
1850
通过率
27%
被复制
9
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