1092. To Buy or Not to Buy (20)

珠串购买问题解析

最新推荐文章于 2021-08-21 20:02:21 发布

转载最新推荐文章于 2021-08-21 20:02:21 发布 · 35 阅读

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原文链接：http://www.cnblogs.com/patforjiuzhou/p/7147246.html

本文介绍了一种算法，用于解决珠子串购问题。通过对比两串珠子的颜色及数量，判断是否能满足需求并计算额外购买或缺失的数量。采用map进行颜色及数量的存储与对比。

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

这道题并不难。注意点：1 使用map存储对比时，还有可能另一个map里没有存储相关数据 2 在这种情况下，ncnt+map中该项的数量，而不是1

#include <map>
#include <iostream>
#include <string>
using namespace std;
map<char, int> a_m, b_m;
int main()
{
	string a, b;
	cin >> a >> b;
	int cnt = 0;
	int ncnt = 0;
	for (auto i = a.begin(); i != a.end(); i++)
	{
		a_m[*i]++;
	}
	for (auto i = b.begin(); i != b.end(); i++)
	{
		b_m[*i]++;
	}
	/*
	for (auto i = b_m.begin(); i!=b_m.end(); i++)
	{
		cout << i->first << " " << i->second << endl;
	}
	cout << "a" << endl;
	for (auto i = a_m.begin(); i!=a_m.end(); i++)
	{
		cout << i->first << " " << i->second << endl;
	}
	*/
	int flag=0;
	for (auto i = b_m.begin(); i != b_m.end(); i++)
	{
		for (auto j = a_m.begin(); j != a_m.end(); j++)
		{
			if (i->first == j->first)
			{
				flag = 1;
				if (i->second <= j->second)
					continue;
				else
				{
					//cout << "b " << i->first << " " << i->second << endl;
					//cout << "a " << j->first << " " << j->second << endl;
					ncnt += (i->second - j->second);

					//cout << "b->first " << i->first << " :" << i->second;
					//cout << " a : " << j->first << " :" << j->second << endl;
				}

			}
			else continue;
		}
		if (flag == 0)
		{
			ncnt+=i->second;
		}
		flag = 0;

	}
	if (ncnt == 0)
	{
		cout << "Yes " << (a.size() - b.size());
	}
	else cout << "No " << ncnt;
	return 0;
}

转载于:https://www.cnblogs.com/patforjiuzhou/p/7147246.html