PAT乙级1014

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:

1. 第一对是相同的大写字母A-G
2. 第二对是相同的数字0-9和A-N
3. 小时和分钟输出宽度为2，不足2位用零填充
4. 不用map也行，可以用ASCII码和字符的对应关系
5. 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 <iostream>
#include <string>
#include <map>
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<char,string> 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<int, int> 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<minLen1;i++){
        if (strs[0][i] == strs[1][i] && isHour(strs[1][i])){
            printf("%02d:", hourMap[strs[1][i]]);
            break;
        }
    }

    // 遍历后两个字符串
    int minLen2 = strs[2].length() < 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：https://blog.csdn.net/supremebuct/article/details/83105861

#include <iostream> //9.16
#include <stdlib.h>
#include <string>
#include <cctype>
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);
}

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作者：@臭咸鱼

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