zoukankan      html  css  js  c++  java
  • poj1092. To Buy or Not to Buy (20)

    1092. To Buy or Not to Buy (20)

    时间限制
    100 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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
    

    提交代码

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<queue>
     6 #include<vector>
     7 #include<cmath>
     8 #include<map>
     9 using namespace std;
    10 map<char,int> beads;
    11 int main(){
    12     //freopen("C:INPUT.txt", "r", stdin);
    13     string s1;
    14     cin>>s1;
    15     int i;
    16     for(i=0;i<s1.length();i++){
    17         beads[s1[i]]++;
    18     }
    19     string s2;
    20     cin>>s2;
    21     int count=0;
    22     for(i=0;i<s2.length();i++){
    23         if(!beads[s2[i]]){
    24             count++;
    25             continue;
    26         }
    27         beads[s2[i]]--;
    28     }
    29     if(count){
    30         printf("No %d
    ",count);
    31     }
    32     else{
    33         printf("Yes %d
    ",s1.length()-s2.length());
    34     }
    35     return 0;
    36 }
  • 相关阅读:
    【字符串】C语言_字符串常量详解
    2138=数据结构实验之图论三:判断可达性
    3363=数据结构实验之图论七:驴友计划
    1916=字符串扩展(JAVA)
    2140=数据结构实验之图论十:判断给定图是否存在合法拓扑序列
    3364=数据结构实验之图论八:欧拉回路
    2138=数据结构实验之图论三:判断可达性
    2271=Eddy的难题(JAVA)
    2246=时间日期格式转换(JAVA)
    2804=数据结构实验之二叉树八:(中序后序)求二叉树的深度
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4762168.html
Copyright © 2011-2022 走看看