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 2:

No 2

 #include <iostream>
#include <string>
using namespace std;
string sell, buy;
int nums[] = { }, res = ;//ASCALL表
int main()
{
cin >> sell >> buy;
for (auto s : sell)
nums[s]++;
for (auto s : buy)
{
if (nums[s] == )
res++;
else
nums[s]--;
}
if (res == )
cout << "Yes" << " " << (sell.size() - buy.size()) << endl;
else
cout << "No" << " " << res << endl;
return ;
}

PAT甲级——A1092 To Buy or Not to Buy【20】的更多相关文章

  1. 【PAT甲级】1116 Come on! Let's C (20分)

    题意: 输入一个正整数N(<=10000),接着依次输入N个学生的ID.输入一个正整数Q,接着询问Q次,每次输入一个学生的ID,如果这个学生的ID不出现在之前的排行榜上输出Are you kid ...

  2. 【PAT甲级】1069 The Black Hole of Numbers (20 分)

    题意: 输入一个四位的正整数N,输出每位数字降序排序后的四位数字减去升序排序后的四位数字等于的四位数字,如果数字全部相同或者结果为6174(黑洞循环数字)则停止. trick: 这道题一反常态的输入的 ...

  3. 【PAT甲级】1065 A+B and C (64bit) (20 分)(大数溢出)

    题意: 输入三个整数A,B,C(long long范围内),输出是否A+B>C. trick: 测试点2包括溢出的数据,判断一下是否溢出即可. AAAAAccepted code: #defin ...

  4. PAT甲级题解(慢慢刷中)

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  5. 【转载】【PAT】PAT甲级题型分类整理

    最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel P ...

  6. PAT甲级题分类汇编——杂项

    本文为PAT甲级分类汇编系列文章. 集合.散列.数学.算法,这几类的题目都比较少,放到一起讲. 题号 标题 分数 大意 类型 1063 Set Similarity 25 集合相似度 集合 1067 ...

  7. PAT甲级题分类汇编——序言

    今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是 ...

  8. PAT甲级题解分类byZlc

    专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d& ...

  9. PAT 甲级真题题解(63-120)

    2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分 ...

随机推荐

  1. vue条形码+二维码

    <template> <section style="height: 100vh;" class="bg"> <div class ...

  2. Eclipse指定JDK版本

    Eclipse有好多版本,同时又分32位和64位,要使用相对应的版本和一样位数的JDK,Eclipse才能正常运行. 对应不上时,Eclipse 甚至不能正常启动.报错:“Failed to load ...

  3. iOS开发系列-NSFileManager

    NSFileManager NSFileManager类主要对文件和目录的操作(删除.修改.移动.复制等等).

  4. What is the difference between HTTP_CLIENT_IP and HTTP_X_FORWARDED_FOR

    What is the difference between HTTP_CLIENT_IP and HTTP_X_FORWARDED_FOR? it is impossible to say. Dif ...

  5. Nginx配置两份日志记录

    nginx配置 版本-1.4.4 --- access_log /alidata/log/nginx/access/wordpress1.log ; access_log /alidata/log/n ...

  6. BeanPostProcessor原理学习

    <Spring源码解析>笔记 1.自定义的BeanPostProcessor @Component public class MyBeanPostProcessor implements ...

  7. 基于vue,通过父组件触发子组件的请求,等请求完毕以后,显示子组件,同时隐藏父组件

    正常情况下,子组件应该尽量减少业务逻辑,而应该将业务逻辑放到父组件里面,从而减少耦合,但是当 我们不得不用到这种情况时,可以采用下面的思路 解决方案 尽量将请求单独作为一个函数(不要将请求放到show ...

  8. QQ聊天机器人 Delphi代码

    QQ聊天机器人     前几日,看到杂志上有一篇关于开发QQ聊天机器人的文章.谈到了对QQ循环发送消息内容,感觉倒也很好玩,于是拿起Delphi开始了我的QQ聊天机器人之路. 首先要明白自己要做什么, ...

  9. ajax跨域获取网站json数据

    由于自己的公司的项目需要调用视频地址 1:当为链接时:直接在播放器用数据库查找的地址 2:当为外部链接时:直接用window.location.href('数据库查找的地址') 3:当为H5链接时:使 ...

  10. Http学习(三)

    HTTP的问题: 通信使用明文,可能会遭到窃听:HTTP本身不具备加密功能,根据TCP/IP协议工作的线路上可能会遭到窃听,即使通信内容已经加密,也会被看到 通信加密:通过SSL(Secure Soc ...