codechef Jewels and Stones 题解

最新推荐文章于 2019-06-08 15:01:59 发布

转载最新推荐文章于 2019-06-08 15:01:59 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/mfrbuaa/p/3768508.html

本文介绍了一种通过对比两个字符串来计数特定字符出现次数的算法。该算法使用哈希表记录宝石字符，并遍历石头字符串来判断是否为宝石，以此实现宝石计数功能。

Soma is a fashionable girl. She absolutely loves shiny stones that she can put on as jewellery accessories. She has been collecting stones since her childhood - now she has become really good with identifying which ones are fake and which ones are not. Her King requested for her help in mining precious stones, so she has told him which all stones are jewels and which are not. Given her description, your task is to count the number of jewel stones.

More formally, you're given a string J composed of latin characters where each character is a jewel. You're also given a string S composed of latin characters where each character is a mined stone. You have to find out how many characters of S are in J as well.

Input

First line contains an integer T denoting the number of test cases. Then follow T test cases. Each test case consists of two lines, each of which contains a string composed of English lower case and upper characters. First of these is the jewel string J and the second one is stone string S. You can assume that 1 <= T <= 100, 1 <= |J|, |S| <= 100

Output

Output for each test case, a single integer, the number of jewels mined.

Example

Input:
4
abc
abcdef
aA
abAZ
aaa
a
what
none

Output:
3
2
1
0

查找字符串的问题。

这里一定要熟悉hash表的运用。常常考的。

还有要懂得推断输入结束的符号 - EOF

#pragma once
#include <stdio.h>

class JewelsandStones
{
public:
	JewelsandStones()
	{
		int T = 0;		
		scanf("%d\n", &T);
		while (T--)
		{
			bool J[256] = {false};
			char c;
			while ((c = getchar()) != '\n' && c != EOF)
			{
				J[c] = true;
			}
			int ans = 0;
			while ((c = getchar()) != '\n' && c != EOF)
			{
				if (J[c]) ans++;
			}
			printf("%d\n", ans);
		}
	}
};

int jewelsandStones()
{
	JewelsandStones jewel;
	return 0;
}

转载于:https://www.cnblogs.com/mfrbuaa/p/3768508.html