1092. To Buy or Not to Buy (20)

PAT-A1092，题目地址：https://www.patest.cn/contests/pat-a-practise/1092
可以理解为比较字符串不同（无序）的题目，需要首先将两个字符串分别遍历一次，分别记录每个字符串中各个字符的数量，然后分别比较每个字符串数量的多少，对于每个字符（珠子），如果“需要”的比“卖”的多，则missing增加，如果“需要”的比卖的“少”，则extra增加。详见代码

#include <cstdio>
#include <cstring>

void check_char(int* count, char ch){
    if(ch >= '0' && ch <= '9'){
        count[ch - '0']++;
    }
    else if(ch >= 'a' && ch <= 'z'){
        count[ch - 'a' + 10]++;
    }
    else if(ch >= 'A' && ch <= 'Z'){
        count[ch - 'A' + 10 + 26]++;
    }
}

int main(){
    char shop[1000]; //卖的珠子
    char mine[1000]; //需要的珠子
    int shop_count[62] = {0}; //记录卖的珠子中每个颜色的数量，0-9，a-z，A-Z
    int mine_count[62] = {0};//记录需要的珠子
    scanf("%s %s", shop, mine);
    int shop_length = strlen(shop);
    int mine_length = strlen(mine);

    for(int i = 0; i < shop_length; i++){
        check_char(shop_count, shop[i]);
    }
    for(int i = 0; i < mine_length; i++){
        check_char(mine_count, mine[i]);
    }
    int extra = 0, missing = 0;
    for(int i = 0; i < 62; i++){
        if(shop_count[i] > mine_count[i]){
            extra += shop_count[i] - mine_count[i];
        }
        else if(shop_count[i] < mine_count[i]){
            missing += mine_count[i] - shop_count[i];
        }
        else{}
    }
    if(missing == 0){
        printf("Yes %d\n", extra);
    }
    else{
        printf("No %d\n", missing);
    }
    return 0;
}