PTA甲级1092 To Buy or Not to Buy (20point(s))

首先,先贴柳神的博客

https://www.liuchuo.net/ 这是地址

想要刷好PTA,强烈推荐柳神的博客,和算法笔记

题目原文

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

生词如下：

bead 弹珠

题目大意

Eva去商店买弹珠,她有若干想买的弹珠颜色,但是商店卖的都是一整串的弹珠,一整串的弹珠可能会多出来,也可能会缺。

现在给你eva要买的弹珠颜色和个数和商店里有的弹珠颜色和个数,

要你输出Yes或No,Yes要输出多了几个,No要输出缺了几个.

代码如下

#include 
using namespace std;
int book[123];		//用一个int来存储所有的数
int main() {
	string a, b;
	cin >> a >> b;
	for (int i = 0; i < a.length(); i++)
		book[a[i]]++;		//就加加
	int result = 0;
	for (int i = 0; i < b.length(); i++) {
		if (book[b[i]] > 0)
			book[b[i]]--;	//有相同的,就减减
		else
			result++;
	}
	if (result != 0)
		printf("No %d", result);
	else
		printf("Yes %d", a.length() - b.length());
	return 0;
}