1092.To Buy or Not to Buy

【题意】

        判断字符串1是否包含了字符串2所有的字符，是的话输出多余字符数，否则输出缺的字符数

【思路】

        直接码即可

#include <iostream>#include <string>#include <map>using namespace std;int main(){string str[2];map<char, int> mm;for (int i = 0; i < 2; ++i){cin >> str[i];}for (auto i : str[0]){++mm[i];}int missCnt, extraCnt;missCnt = extraCnt = 0;for (auto i : str[1]){if (mm.find(i) != mm.end() && mm[i]){--mm[i];}else{++missCnt;}}if (missCnt){cout << "No " << missCnt;}else{for (auto i : mm){extraCnt += i.second;}cout << "Yes " << extraCnt;}system("pause");return 0;}