zoukankan      html  css  js  c++  java
  • ZOJ 3207 80ers' Memory(水题)

    80ers' Memory

    Time Limit: 1 Second      Memory Limit: 32768 KB

    I guess most of us are so called 80ers, which means that we were born in the 1980's. This group of people shared a lot of common memories. For example, the Saint Seiya, the YoYo ball, the Super Mario, and so on. Do you still remember these?

    Input

    There will be ONLY ONE test case.

    The test case begins with a positive integer N, (N < 100).
    Then there will be N lines each containing a single word describing a keyword of the typical 80ers' memories. Each word consists of letters, [a-zA-Z], numbers, [0-9], and the underline, '_'. The length of each word would be no more than 20.
    Then one line follows with a positive integer K, (K < 100).
    The last part of the test case will be K lines. The i-th line contains the keywords given by the i-th person and starts with a positive integer Ni. Then there will be Ni words separated by white spaces. Each of these words has no more than 20 characters.
    All the words are case sensitive.

    Output

    For each of these K people, you are asked to count the number of typical 80ers' keywords on his/her list and output it in a single line.

    Sample Input

    4
    Saint_Seiya
    YoYo_ball
    Super_Mario
    HuLuWa
    3
    2 Saint_Seiya TiaoFangZi
    1 KTV
    3 HuLuWa YOYO_BALL YoYo_ball
    

    Sample Output

    1
    0
    2
    

    Author: GAO, Fei
    Source: The 6th Zhejiang Provincial Collegiate Programming Contest

    map秒过了。。。

    水题。。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<algorithm>
    #include<map>
    using namespace std;
    map<string,int>mp;
    int main()
    {
        int n,k;
        int m;
        string str;
        while(scanf("%d",&n)!=EOF)
        {
            mp.clear();
            while(n--)
            {
                cin>>str;
                if(mp[str]==0)mp[str]=1;
            }
            int ans;
            scanf("%d",&k);
            while(k--)
            {
                scanf("%d",&m);
                ans=0;
                while(m--)
                {
                    cin>>str;
                    ans+=mp[str];
                }
                printf("%d\n",ans);
            }
        }
        return 0;
    }
  • 相关阅读:
    2019.6.30 Spring注解 bean后置处理器和属性赋值
    2019.6.29 Spring注解-Bean的注册
    2019.6.28 Spring注解
    boost基础环境搭建
    动态规划入门一:钢条切割问题
    《剑指offer》读书笔记
    字符串的全排列
    西山居递归面试题
    常见的数据结构
    832. Flipping an Image
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2706130.html
Copyright © 2011-2022 走看看