pat 1039 in the end buying

I want to buy some red beads made a bunch of their favorite beads. There are many stall selling beads strings of colorful beads, but refused to sell any string split up. So you want to help judge a little red 
, the house contains a beaded beads all they want? If so, tell her how much extra beads; if not, then tell her how much missing beads. 

For your convenience, we use [0-9], [az], the characters within [AZ] to represent the range of colors. For example, YrR8RrY want a little red beads; ppRYYGrrYBR2258 can buy it, because it contains 
all the beads she wants, does not need more than the 8 beads; ppRYYGrrYB225 can not buy, because there is no black beads, and a less pieces of red beads.

 

Enter a description:

Each input comprises a test. Each test case separately in two lines has given beads and red beads wanted to stall, no more than 1000 two strings of beads.

Output Description:

If you can buy, the output is "Yes" and how many extra beads in a row; if not buy, the output in a row "No" and the number of missing beads. Meanwhile separated by a space.

 

Input example:

ppRYYGrrYBR2258
YrR8RrY

 

Output example:

Yes 8 
　　idea: to process the string. string.erase () code is as follows:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
int main()
{
	string a;
	string b;
	cin>>a>>b;
	bool flag=true;
	int len1=a.length();
	int len2=b.length();
	int count=0;
	for(int i=0;i<len2;i++)
	{
		int temp=a.find(b[i],0);//不断更新下标，当找不到时，计数
		if(temp!=-1)
		{
			a.erase(a.begin()+temp);
		}
		else {
			flag=false;
			count++;
		}
	}
	if(flag)
	{
		printf ( "Yes% d \ n ", len1-len2); // time to buy, can be calculated by comparing the length of the string number string plurality 
	} 
	the else the printf ( "% D No \ n-", COUNT); 
	return 0; 
}