Pat(A) 1092. To Buy or Not to Buy (20)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1092

1092. To Buy or Not to Buy (20)

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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

题目大意

给定两个串，看第一个串是否包含第二个串中所有的字符（包括个数），如果全部包含，计算多余的字符数，如果不全包含，计算所缺的字符数。

解题报告

把字符对应成下标，统计第一个串每个字符的个数，统计第二个串每个字符的个数，然后进行比较

代码

#include "iostream"#include "map"#include "string"using namespace std;map<char,int> c2i;int eva[65];int shop[65];int more=0,miss=0;void init(){    string a,b;    cin>>a>>b;    int i,id;    for(i = '0'; i <= '9'; i++){ // 此时i可理解为char型        id = i - '0';        c2i[i] = id;    }    for(i = 'a'; i <= 'z'; i++){ // 此时i可理解为char型        id = i - 'a' + 10;        c2i[i] = id;    }    for(i = 'A'; i <= 'Z'; i++){ // 此时i可理解为char型        id = i - 'A' + 36;        c2i[i] = id;    }    for(i = 0; i < a.length(); i++){        shop[c2i[a[i]]] ++;    }    for(i = 0; i < b.length(); i++){        eva[c2i[b[i]]] ++;    }}void cal(){    for(int i= 0; i < 62; i++){        if(shop[i] > eva[i])            more += shop[i] - eva[i];        else            miss += eva[i] - shop[i];    }}int main(){    init();    //prin();    cal();    if(miss > 0)        printf("No %d",miss);    else        printf("Yes %d",more);    //system("pause");}