http://www.patest.cn/contests/pat-a-practise/1092

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

这道题是2015考研机试前的那个PAT的D题 http://www.patest.cn/contests/pat-a-101-125-1-2015-03-14

这道题很简单,所以考试结束后原英文题目放到了A中,中文版的也放到了B中    http://www.cnblogs.com/asinlzm/p/4441575.html

 #include<cstdio>
#include<cstring> int main()
{
char str[];
gets(str); int istr=,charnum[]={};
while(str[istr]) charnum[str[istr]]++,istr++; // count beads which belong to the shop owner gets(str);
istr=;
while(str[istr]) charnum[str[istr]]--,istr++; // count beads which Eva need int more=,less=;
for(int i=;i<;i++)
{
if(charnum[i]>) more+=charnum[i];
else if(charnum[i]<) less-=charnum[i];
} if(less) printf("No %d",less);
else printf("Yes %d",more);
return ;
}

最新文章

  1. 学习Maven之Maven Clean Plugin
  2. 四校联考 tree3
  3. js添加var和不加var区别
  4. 评价网站UI--北京邮电大学首页
  5. Tomcat 服务器版本的区别以及下载与安装
  6. Hadoop学习笔记: HDFS
  7. C++学习33 函数模板
  8. 《Java数据结构与算法》笔记-CH5-链表-4用链表实现堆栈
  9. poj 1742 Coins
  10. MAC 终端 显示隐藏文件 关闭显示隐藏文件
  11. kiss框架学习
  12. react-native android textinput显示不全的问题
  13. C#获取文件夹下面的所有文件名
  14. 三台机器之间root用户ssh互信配置
  15. (zhuan) 资源|TensorFlow初学者必须了解的55个经典案例
  16. Android - AMS源码分析
  17. 【小白的CFD之旅】20 计算区域的构建
  18. PAT 1072 开学寄语
  19. Linux中如何安装Apache服务器
  20. 基于Qt5 跨平台应用开发

热门文章

  1. thinkphp5缓存使用
  2. jmeter csv中获取带引号的数据详情(转)
  3. BZOJ 4165 矩阵 堆
  4. 11-散列4 Hashing - Hard Version (30 分)
  5. DbUtils(一) 结果集概览
  6. 09.Spring Bean 注册 - BeanDefinitionRegistry
  7. A -- A. Quailty and Playing Cards 模拟 + 思考
  8. map 常用方法
  9. 机器学习框架ML.NET学习笔记【2】入门之二元分类
  10. 《C#高效编程》读书笔记12-使用推荐成员初始化器而不是赋值语句