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

 #include<cstdio>

 #include<iostream>

 using namespace std;

 int trans(char c){

     if(c >= '' && c <= '')

         return c - '';

     else if(c >= 'a' && c <= 'z')

         return c - 'a' + ;

     else return c - 'A' + ;

 }

 int main(){

     char shop[], Eva[];

     int hashTB[] = {,};

     int flag = ;

     scanf("%s %s", shop, Eva);

     for(int i = ; shop[i] != '\0'; i++)

         hashTB[trans(shop[i])]++;

     for(int i = ; Eva[i] != '\0'; i++){

         hashTB[trans(Eva[i])]--;

         if(hashTB[trans(Eva[i])] < )

             flag = ;

     }

     int ans = ;

     if(flag == ){

         for(int i = ; i < ; i++)

             if(hashTB[i] < )

                 ans += hashTB[i];

         printf("No %d", ans * -);

     }else{

         for(int i = ; i < ; i++)

             ans += hashTB[i];

         printf("Yes %d", ans);

     }

     cin >> Eva;

     return ;

 }

总结：

1、本题大意：比较店家和EVa的珠子，如果店家的珠子比Eva所需的还要多，那就输出多的个数。如果店家缺少Eva所需的某几种珠子，那么输出所缺的个数。

2、如果有多余的珠子，可以直接输出两串珠子的长度差，而不必统计累加。