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

最新推荐文章于 2025-10-11 12:44:10 发布

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本文介绍了一种算法，用于解决珠子串购买问题。具体来说，算法检查商店拥有的珠子串是否包含客户所需的所有珠子。如果包含，则计算额外需要购买的珠子数量；如果不包含，则计算缺少的珠子数量。文章提供了AC代码实现。

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.
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
题目大意：第一行给出商店拥有的珠子，第二行给出你需要的珠子；
计算是否商店拥有你所需的所有珠子若是则输出Yes，以及需要额外购买的珠子（英文商店珠子只能一串一串买）；否则输出No,并输出缺少的珠子；
分析：标准的散列问题；
AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
 int HASH[128]={0};
int main(){
char a[1001],b[1001];
bool flag=true;
scanf("%s",a);
getchar();
scanf("%s",b);
int len1=strlen(a), len2=strlen(b),count=0;
for(int i=0;i<len1;i++)
   HASH[a[i]]++;
for(int i=0;i<len2;i++){
if(HASH[b[i]]!=0)  HASH[b[i]]--;
else  flag=false,count++;
}
if(flag)
    printf("Yes %d",len1-len2);
else
    printf("No %d",count);
system("pause");
return 0;
}