题目链接

https://pintia.cn/problem-sets/994805260223102976/problems/994805308755394560

题解一（部分正确）

这是我的方法，第2个测试点没有过，和正确的代码比较，目前没比较出来错误，可能是我map用错了？

需要注意的点:

第一对是相同的大写字母A-G

第二对是相同的数字0-9和A-N

小时和分钟输出宽度为2，不足2位用零填充

不用map也行，可以用ASCII码和字符的对应关系

判断大小写字母、数字等函数C++已自带，不用自己写

// PAT BasicLevel 1014// https://pintia.cn/problem-sets/994805260223102976/problems/994805308755394560#include 
    
     #include 
     
      #include 
      using namespace std;bool isUpperCase(char);bool isLowerCase(char);bool isAlpha(char);bool isNumber(char);bool isDay(char);bool isHour(char);int main(){    // 获取四个字符串    string strs[4];    for(int i=0;i<4;++i){        cin >> strs[i];    }    // 字母与周几的映射    map
       
         dayMap;    dayMap['A'] = "MON";dayMap['B'] = "TUE";dayMap['C'] = "WED";    dayMap['D'] = "THU";dayMap['E'] = "FRI";dayMap['F'] = "SAT";dayMap['G'] = "SUN";    // 字母（或数字）与小时的映射    map
        
          hourMap;    hourMap[0] = 0;hourMap[1] = 1;hourMap[2] = 2;hourMap[3] = 3;hourMap[4] = 4;    hourMap[5] = 5;hourMap[6] = 6;hourMap[7] = 7;hourMap[8] = 8;hourMap[9] = 9;    hourMap['A'] = 10;hourMap['B'] = 11;hourMap['C'] = 12;hourMap['D'] = 13;    hourMap['E'] = 14;hourMap['F'] = 15;hourMap['G'] = 16;hourMap['H'] = 17;    hourMap['I'] = 18;hourMap['J'] = 19;hourMap['K'] = 20;hourMap['L'] = 21;    hourMap['M'] = 22;hourMap['N'] = 23;    // 遍历前两个字符串    int index;    int minLen1 = strs[0].length() < strs[1].length() ? strs[0].length() : strs[1].length();    for (int i = 0; i < minLen1; ++i){        if (strs[0][i] == strs[1][i] && isDay(strs[1][i])){            cout << dayMap[strs[1][i]] << ' ';            index=i;            break;        }    }    for(int i=index+1;i
         
          < strs[3].length() ? strs[2].length() : strs[3].length(); for (int i = 0; i < minLen2; ++i){ if (strs[2][i] == strs[3][i] && isAlpha(strs[3][i])){ printf("%02d", i); break; } } //system("pause"); return 0;}bool isDay(char c){ // A-G return c >= 'A' && c <= 'G';}bool isHour(char c){ // 0-9 A-N return isNumber(c) || (c >= 'A' && c <= 'N');}bool isUpperCase(char c){ // A-Z return c >= 'A' && c <= 'Z';}bool isLowerCase(char c){ // a-z return c >= 'a' && c <= 'z';}bool isAlpha(char c){ // a-z A-Z return isLowerCase(c)||isUpperCase(c);}bool isNumber(char c){ // 0-9 return c >= '0' && c <= '9';}

题解二

网上找的，和我看起来思路一样啊……

参考链接：

#include 
    
      //9.16#include 
     
      #include 
      
       #include 
       
        using namespace std;void deal(string ch1, string ch2);void deal1(string ch3, string ch4);int main(){    string ch1, ch2, ch3, ch4;    cin >> ch1 >> ch2 >> ch3 >> ch4;    deal(ch1, ch2);    deal1(ch3, ch4);    return 0;}void deal(string ch1, string ch2){    string day[] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};    int num, num1, i = 0;    while (i < ch1.length() && i < ch2.length())    {        if (ch1[i] == ch2[i] && (ch1[i] >= 'A' && ch1[i] <= 'G')) //一开始没有设定范围只是判断了其是否是大写，导致有的例子通不过        {            num1 = ch1[i] - 'A';            break;        }        i++;    }    i++;    cout << day[num1] << ' ';    int num2;    while (i < ch1.length() && i < ch2.length())    {        if (ch1[i] == ch2[i])        {            if (isdigit(ch1[i]))            {                num2 = ch1[i] - '0';                break;            }            else if (ch1[i] >= 'A' && ch1[i] <= 'N') //一开始没有设定范围只是判断了其是否是大写，导致有的例子通不过            {                num2 = 10 + (ch1[i] - 'A');                break;            }        }        i++;    }    printf("%02d:", num2);}void deal1(string ch3, string ch4){    int i = 0;    int num3;    while (i < ch3.length() && i < ch4.length())    {        if (ch3[i] == ch4[i] && isalpha(ch3[i]))        {            num3 = i;            break;        }        i++;    }    printf("%02d", num3);}