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题目

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

题意

字符串a,b

• b中字符出现次数<=其在a中出现次数，输出Yes a中多余字符出现次数
• b中字符出现次数>其在a中出现次数，输出No a中缺少字符数

题目分析

算法1

1. 统计a中字符出现的次数，记录在asc数组中
2. 使用df记录缺少字符数
3. 遍历b中字符，当前字符为b[i]
   • 若asc[b[i]]大于0，减一
   • 若asc[b[i]]等于0，df++（缺少数+1）
4. 判断df值，并打印
   • 若df==0，表明不缺少字符，输出a中多余字符--a的长度-b的长度
   • 若df!=0，表明缺少字符，输出df

算法2

1. 统计a,b中字符出现次数，记录在容器asc1,asc2中
2. 使用df记录缺少字符数
3. 遍历b
   • 若asc1[b[i]]<asc2[b[i]]，df++（缺少数+1）;
   • 若asc2[b[i]]>=asc2[b[i]]，不缺少，跳过
4. 判断df值，并打印
   • 若df==0，表明不缺少字符，输出a中多余字符--a的长度-b的长度
   • 若df!=0，表明缺少字符，输出df

Code

Code 01（算法1、最优）

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char * argv[]){
    string a,b;
    cin>>a>>b;
    int asc[256]={0};
    for(int i=0;i<a.length();i++){
        asc[a[i]]++;
    } 
    int df=0;
    for(int i=0;i<b.length();i++){
        if(asc[b[i]]>0)asc[b[i]]--;
        else df++; 
    }
    if(df==0)printf("Yes %d",a.length()-b.length());
    if(df!=0)printf("No %d",df);
    return 0;
}

Code 02（算法2、int array）

#include <iostream>
#include <cstring>
using namespace std;
int main(int argc, char * argv[]) {
    char s1[1001];
    char s2[1001];
    cin.getline(s1,1001);
    cin.getline(s2,1001);
    int len1=strlen(s1),len2=strlen(s2);
    int asc1[256]= {0};
    int asc2[256]= {0};
    for(int i=0; i<len1; i++) asc1[s1[i]]++;
    for(int i=0; i<len2; i++) asc2[s2[i]]++;
    int df=0;
    int ascp[256]= {0};
    for(int i=0; i<len2; i++) {
        if(ascp[s2[i]]==0&&asc2[s2[i]]>asc1[s2[i]]) {
            df+=asc2[s2[i]]-asc1[s2[i]];
            ascp[s2[i]]=1;
        }
    }
    if(df==0)printf("Yes %d",len1-len2);
    if(df!=0)printf("No %d",df);
    return 0;
}

Code 03（算法2、map）

#include <iostream>
#include <cstring>
#include <unordered_map>
using namespace std;
int main(int argc, char * argv[]) {
    char s1[1001];
    char s2[1001];
    cin.getline(s1,1001);
    cin.getline(s2,1001);
    int len1=strlen(s1),len2=strlen(s2);
    unordered_map<char,int> m1,m2;
    for(int i=0; i<len1; i++) m1[s1[i]]++;
    for(int i=0; i<len2; i++) m2[s2[i]]++;
    int df=0;
    int ascp[256]= {0};
    for(int i=0; i<len2; i++) {
        if(ascp[s2[i]]==0&&m2[s2[i]]>m1[s2[i]]) {
            df+=m2[s2[i]]-m1[s2[i]];
            ascp[s2[i]]=1;
        }
    }
    if(df==0)printf("Yes %d",len1-len2);
    if(df!=0)printf("No %d",df);
    return 0;
}

PAT Advanced 1092 To Buy or Not to Buy (20) [Hash散列]

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原文地址：https://www.cnblogs.com/houzm/p/12239292.html