1092. To Buy or Not to Buy (20)

来源：互联网发布：最火的中文编程编辑：程序博客网时间：2024/05/01 14:37

#include <stdio.h>#include <string.h>using namespace std;int main(){    char ch[1001];    char ch1[1001];    gets(ch);    gets(ch1);    int index[101];    for(int i=0;i<101;i++)    {        index[i]=0;    }    for(int i=0;i<strlen(ch);i++)    {        if(ch[i]>='0'&&ch[i]<='9')            index[ch[i]-'0']++;        if(ch[i]>='a'&&ch[i]<='z')            index[ch[i]-'a'+10]++;        if(ch[i]>='A'&&ch[i]<='Z')            index[ch[i]-'A'+40]++;    }    for(int i=0;i<strlen(ch1);i++)    {        if(ch1[i]>='0'&&ch1[i]<='9')            index[ch1[i]-'0']--;        if(ch1[i]>='a'&&ch1[i]<='z')            index[ch1[i]-'a'+10]--;        if(ch1[i]>='A'&&ch1[i]<='Z')            index[ch1[i]-'A'+40]--;    }    int count=0;    int count2=0;    int flag=1;    for(int i=0;i<101;i++)    {        if(index[i]>0)            count+=index[i];        if(index[i]<0)        {            flag=0;            count2-=index[i];        }    }    if(flag)        printf("Yes %d",count);    else        printf("No %d",count2);}

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