PAT A 1092 B 1039（甲级 乙级）

1092 To Buy or Not to Buy (20)（20 分）

作者: CHEN, Yue

单位: PAT联盟

时间限制: 100ms

内存限制: 64MB

代码长度限制: 16KB

Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must check whether a string in the shop contains all the beads she needs. She now comes to you for help: if the answer is "Yes", please tell her the number of extra beads she has to buy; or if the answer is "No", please tell her the number of beads missing from the string.

For the sake of simplicity, let's use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. For example, the 3rd string in Figure 1 is the one that Eva would like to make. Then the 1st string is okay since it contains all the necessary beads with 8 extra ones; yet the 2nd one is not since there is no black bead and one less red bead.

\ Figure 1

Input Specification:

Each input file contains one test case. Each case gives in two lines the strings of no more than 1000 beads which belong to the shop owner and Eva, respectively.

Output Specification:

For each test case, print your answer in one line. If the answer is "Yes", then also output the number of extra beads Eva has to buy; or if the answer is "No", then also output the number of beads missing from the string. There must be exactly 1 space between the answer and the number.

Sample Input 1:

ppRYYGrrYBR2258
YrR8RrY

Sample Output 1:

Yes 8

Sample Input 2:

ppRYYGrrYB225
YrR8RrY

Sample Output 1:

No 2

 

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首先很明显可以看出要用散列表的思想去解决这个问题，然后又由于数据范围比较小，可以直接是用一个int数组将字符转换为下标进行计数即可。

 

在《算法笔记》给出的解法中，是先对商店珠子统计然后再计算需要的珠子，如果数量小于0了就开始增加miss变量的值，这样在输出缺多少个珠子的时候非常方便，而我的代码中是先统计需要后计算商店的珠子是否达标，然后再遍历一边数组去查看是否有负值来决定是否达标，在这一方面感觉还是书中给出的代码更优秀一些。

 

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AC代码：

#include 
#include 

int main() {
    int c[200] = {0};
    char shop[1010], need[1010];

    scanf("%s", shop);
    scanf("%s", need);

    for(int i = 0; need[i] != '\0'; ++i) {
        ++c[(int)need[i]];
    }

    for(int i = 0; shop[i] != '\0'; ++i) {
        --c[(int)shop[i]];
    }

    int total = 0;
    bool success = true;
    for(int i = 0; i < 200; ++i) {
        if(c[i] > 0) {
            success = false;
            total += c[i];
        }
    }

    if(success) {
        printf("Yes %d\n", strlen(shop) - strlen(need));
    } else {
        printf("No %d\n", total);
    }

    return 0;
}

 

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如有错误，欢迎指摘。