poj1092. To buy or don't to buy (20)

1092. To buy or not to buy (20) time limit MS Memory limit 65536 KB code length limit 16000 B procedure StandardAuthor Chen, Yue

Eva would 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 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 help:if the answer are "Yes", please tell her the number of the extra beads she had 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 of that Eva would like to make. Then the 1st string was okay since it contains all the necessary beads with 8 extra ones; Yet the 2nd one is not since there are no black bead and one less red bead.

Figure 1

Input Specification:

Each input file contains the one test case. Each case gives in lines the strings of no more than the beads which belong to the shop owner and Eva, respectively.

Output Specification:

For each test case, print your answer on one line. If the answer is "Yes" and then also output the number of extra beads Eva have to buy; Or if the answer is "No" and 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:

Ppryygrrybr2258yrr8rry

Sample Output 1:

Yes 8

Sample Input 2:

Ppryygrryb225yrr8rry

Sample Output 1:

No 2

Submit Code

1#include <cstdio>2#include <algorithm>3#include <iostream>4#include <cstring>5#include <queue>6#include <vector>7#include <cmath>8#include <map>9 using namespacestd;Tenmap<Char,int>beads; One intMain () { A     //freopen ("C:input.txt", "R", stdin); -     stringS1; -Cin>>S1; the     inti; -      for(i=0; I<s1.length (); i++){ -beads[s1[i]]++; -     } +     stringS2; -Cin>>S2; +     intCount=0; A      for(i=0; I<s2.length (); i++){ at         if(!Beads[s2[i]]) { -count++; -             Continue; -         } -beads[s2[i]]--; -     } in     if(count) { -printf"No%d\n", count); to     } +     Else{ -printf"Yes%d\n", S1.length ()-s2.length ()); the     } *     return 0; $}

poj1092. To buy or don't to buy (20)