## Pat(A) 1092. To Buy or Not to Buy (20)

### 1092. To Buy or Not to Buy (20)

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

### 代码

#include "iostream"
#include "map"
#include "string"
using namespace std;

map<char,int> c2i;

int eva;
int shop;
int more=0,miss=0;

void init(){
string a,b;
cin>>a>>b;
int i,id;
for(i = '0'; i <= '9'; i++){ // 此时i可理解为char型
id = i - '0';
c2i[i] = id;
}
for(i = 'a'; i <= 'z'; i++){ // 此时i可理解为char型
id = i - 'a' + 10;
c2i[i] = id;
}
for(i = 'A'; i <= 'Z'; i++){ // 此时i可理解为char型
id = i - 'A' + 36;
c2i[i] = id;
}
for(i = 0; i < a.length(); i++){
shop[c2i[a[i]]] ++;
}
for(i = 0; i < b.length(); i++){
eva[c2i[b[i]]] ++;
}
}

void cal(){
for(int i= 0; i < 62; i++){
if(shop[i] > eva[i])
more += shop[i] - eva[i];
else
miss += eva[i] - shop[i];
}
}

int main(){
init();
//prin();
cal();
if(miss > 0)
printf("No %d",miss);
else
printf("Yes %d",more);
//system("pause");
}

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