1092 To Buy or Not to Buy (20 分) PAT 甲级 1092 To Buy or Not to Buy (20 分)

缺乏、安全感 2024-02-19 21:38 38阅读 0赞

## 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.

![figbuy.jpg][]

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

思路：

用了

map<char,int>

统计商品里每个字符和所需字符。

遍历所需字符，如果商品里的对应字符数多于所需字符数，减去。

最后看所需字符的映射是否为空，空，则yes，用商品的长度减去所需字符的长度。否之no 输出缺几个。

#include<iostream>
    #include<map>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	string a,b;
    	int i,j;
    	map<char,int> aa,bb;
    	cin>>a>>b;
    	for(i=0;i<a.length();i++)
    	if(aa.count(a[i])==0)
    	aa.insert(make_pair(a[i],1));
    	else
    	aa[a[i]]++;
    	for(i=0;i<b.length();i++)
    	if(bb.count(b[i])==0)
    	bb.insert(make_pair(b[i],1));
    	else
    	bb[b[i]]++;
    //	cout<<aa.size()<<"   "<<bb.size()<<endl;
    	for(map<char,int>::iterator it=bb.begin();it!=bb.end();)
    	{
    		if(aa.find(it->first)!=aa.end())
    		{
    			if(aa[it->first]>=it->second)
    			{
    				map<char,int>::iterator ir=it;it++;
    				bb.erase(bb.find(ir->first));
    				
    			}
    			else
    			{
    				bb[it->first]-=aa[it->first];
    				it++;
    			}
    		}
    		else
    		it++;
    	}
    //	cout<<1<<endl;
    	if(bb.size()!=0)
    	{
    		int cnt=0;
    		for(map<char,int>::iterator id=bb.begin();id!=bb.end();id++)
    		cnt+=id->second;
    		cout<<"No "<<cnt<<endl;
    	}
    	else
    	{
    		cout<<"Yes "<<a.length()-b.length()<<endl;
    	}
    
    	
    	return 0;
    }

[figbuy.jpg]: https://image.dandelioncloud.cn/pgy_files/images/2024/01/29/4592df2dd2064534a817fbbbbb1b0191.png