PAT 1014 福尔摩斯的约会

来源：互联网发布：非线性最优化方法编辑：程序博客网时间：2024/05/16 15:09

char week_table[7][3]= {“MON”, “TUE”, “WED”, “THU”, “FRI”, “SAT”, “SUN”};就行了定义成二维数组存储时可以简单看作一排一个字符串数组; const char *week_table[7]= {“MON”, “TUE”, “WED”, “THU”, “FRI”, “SAT”, “SUN”};这样是定义了一个指针数组指针元素分别指向每一个对应的字符串存储的首地址

#include<stdio.h>#include <string.h>int main(void){  const char  *week_table[7]= {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};    char str1[61], str2[61], str3[61], str4[61];    int i;    int week, hour, second;    int len1, len2, len3, len4;    scanf("%s%s%s%s", str1, str2, str3, str4);    len1 = strlen(str1);    len2 = strlen(str2);    len3 = strlen(str3);    len4 = strlen(str4);    week = hour = second = 0;    for(i = 0; i < len1 && i < len2; i++)    {        if(str1[i] == str2[i] && (str1[i] >= 'A' && str1[i] <= 'G'))        {            week = str1[i] - 'A';             break;        }    }    for(i++; i < len1 && i < len2; i++)    {        if(str1[i] == str2[i] && ((str1[i] >= 'A' && str1[i] <= 'N') || (str1[i] >= '0' && str1[i] <= '9')))        {            if(str1[i] >= 'A' && str1[i] <= 'N')                hour = str1[i] - 'A' + 10;            else                hour = str1[i] - '0';            break;        }    }    for(i = 0; i < len3 && i < len4; i++)    {        if((str3[i] == str4[i]) && ((str3[i] >= 'a' && str3[i] <= 'z') || (str3[i] >= 'A' && str3[i] <= 'Z')))        {            second = i;            break;        }    }    printf("%s %02d:%02d", week_table[week], hour, second);//2是宽度,如果整数不够2列就补上0    return 0;}

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