【PAT】1092 To Buy or Not to Buy (20分)---简单字符（串）hash

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#字符串 #哈希

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本文介绍了一种用于帮助用户决定是否在商店购买珠串的算法。通过比较商店中珠串的颜色和用户所需颜色，算法能判断出珠串是否包含所有需要的颜色，并计算出额外需要购买的珠子数量或缺失的珠子数量。

1092 To Buy or Not to Buy (20分)

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.
在这里插入图片描述
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

思路总结

为简单哈希表，把数字、大小写字母按ASCII值对应到hash表再按要求查询即可。

AC代码

#include <cstdio>
#include <cstring>
const int maxn = 130;
int hash[maxn] = {0};
char all[1010];
char str[1010];
int main()
{
	bool tag = true; //标志位，表示要买的色号是否齐全
	int less = 0;  //表示少几个
	scanf("%s%s",all,str);
	for(int i=0; i<strlen(all); i++){
		int key = all[i] - '\0';
		hash[key]++;
	}

	for(int i=0; i<strlen(str); i++){
		char c = str[i];
		int key = c - '\0';
		if(hash[key] == 0 ){
			tag = false;
			less ++;
		}
		else{
			hash[key]--;
		}
	}
	if(tag){
		int sum = 0; 	//多出的
		for(int i=0; i<maxn; i++){
			sum += hash[i];
		}
		printf("Yes %d\n",sum );

	}
	else{
		printf("No %d\n",less );
	}
	return 0;
}