1092 To Buy or Not to Buy (20 分)

1092 To Buy or Not to Buy (20 分)

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 2:

No 2

简单题(都没用到hash)，复习string  API的使用  

string.erase
1.删除单个元素
    str.erase(it) 删除迭代器it所指位置的元素
2.删除区间元素
    str.erase(first,end); 删除迭代器区间[first,end)的所有元素
    str.erase(index,length); 删除index下标开始，后面数共length个字符? ★★

string.find('a')

#include<bits/stdc++.h>
using namespace std;
int main(){
  //freopen("in.txt","r",stdin);
  string str1,str2;
  bool flag=true;
  int less=0;
  cin>>str1>>str2;
  for(int i=0;i<str2.length();i++){
    int n=str1.find(str2[i]);//返回的是下标   string::npos转换为int时本质是-1 表示未找到  输出验证过了 
    if(n==string::npos){
      less++;
      flag=false;
    }else{//找到了 用过了  删除 
      str1.erase(n,1);//下标n开始 删除一个字符    
    }
  }
  
  if(flag){
    cout<<"Yes "<<str1.length(); 
  } else{
    cout<<"No "<<less;
  }
  return 0;
}