1092 To Buy or Not to Buy（pass简单方法）

最新推荐文章于 2024-05-24 14:28:31 发布

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最新推荐文章于 2024-05-24 14:28:31 发布

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本文介绍了一个帮助用户Eva选择合适珠串的算法。Eva希望用喜爱的颜色制作珠串，但商店只出售整条珠串。算法通过比较Eva的需求与现有珠串，判断是否能满足需求，并计算额外购买或缺失的数量。

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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 2:

No 2

#include<iostream>
#include<string>
#include<map>
#include<math.h>
using namespace std;



int main() {
	//while (1) {	
		char c;
		map<char,int> s,s1;
		int x,sum0=0,sum1=0;
		map<char,int>::iterator li;
		while((c=getchar())!='\n'){
			++sum0;
			//cout<<c;
			if(s.find(c)==s.end()){
				s[c]=1;
			}else{
				++s[c];
			}
			//cout<<"   "<<s[c]<<endl;
		}		
		while((c=getchar())!='\n'){
			++sum1;
			//cout<<c;
			if(s1.find(c)==s1.end()){
				s1[c]=1;
			}else{
				++s1[c];
			}
			//cout<<"   "<<s1[c]<<endl;
		}
		//cout<<"sum0="<<sum0<<"  sum1="<<sum1<<endl;

		int que=0,add=0;
		for(li=s1.begin();li!=s1.end();++li){
			if(s.find(li->first)==s.end()){				
				que+=li->second;	
				//cout<<li->first<<que<<" "<<li->second<<endl;
				//这里有一点很奇怪，遇到数字的时候s[li->first]和li->second不相等？？？
			}else if(s[li->first]<s1[li->first]){
				que+=s1[li->first]-s[li->first];			
			}		
		}

		if(que>0){
			cout<<"No "<<que<<endl;		
		}else
			cout<<"Yes "<<sum0-sum1<<endl;

	//}
		
}