PAT-PAT (Advanced Level) Practise To Buy or Not to Buy（20） 【一星级】

题目链接：http://www.patest.cn/contests/pat-a-practise/1092

题面：

1092. To Buy or Not to Buy (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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：

    能否用a串凑出b串，可以输出Yes和多的珠子数，不可以输出No和少的珠子数。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <string>
using namespace std;
#define LL long long 
int cnt[150];
int main()
{
	int len1,len2,ans;
    string a,b;
    cin>>a>>b;
	memset(cnt,0,sizeof(cnt));
	len1=b.length();
	for(int i=0;i<len1;i++)
		cnt[b[i]]++;
	len2=a.length();
	ans=0;
	for(int i=0;i<len2;i++)
	   //如果需要该珠子，需要数-1，并且未被替代珠子数减一 
	   if(cnt[a[i]])
	   {
   		  cnt[a[i]]--;
   		  len1--;
	   }
	   //不需要，即为浪费 
	   else
	      ans++;
    //如果都能被替代 
   	if(len1==0)cout<<"Yes "<<ans<<endl;
   	//不能完全替代 
	else cout<<"No "<<len1<<endl;	  
	return 0;
}