zoukankan      html  css  js  c++  java
  • UVa10474

     Where is the Marble?
    Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

    Description

    Raju and Meena love to play with Marbles. They have got a lot of marbles with numbers written on them. At the beginning, Raju would place the marbles one after another in ascending order of the numbers written on them. Then Meena would ask Raju to find the first marble with a certain number. She would count 1...2...3. Raju gets one point for correct answer, and Meena gets the point if Raju fails. After some fixed number of trials the game ends and the player with maximum points wins. Today it's your chance to play as Raju. Being the smart kid, you'd be taking the favor of a computer. But don't underestimate Meena, she had written a program to keep track how much time you're taking to give all the answers. So now you have to write a program, which will help you in your role as Raju.

    Input

    There can be multiple test cases. Total no of test cases is less than 65. Each test case consists begins with 2 integers: N the number of marbles and Q the number of queries Mina would make. The next N lines would contain the numbers written on the N marbles. These marble numbers will not come in any particular order. Following Q lines will have Q queries. Be assured, none of the input numbers are greater than 10000 and none of them are negative.

    Input is terminated by a test case where N = 0 and Q = 0.

    Output

    For each test case output the serial number of the case.

    For each of the queries, print one line of output. The format of this line will depend upon whether or not the query number is written upon any of the marbles. The two different formats are described below:

    • `x found at y', if the first marble with number x was found at position y. Positions are numbered 1, 2,..., N.
    • `x not found', if the marble with number x is not present.

    Look at the output for sample input for details.

    Sample Input

    4 1
    2
    3
    5
    1
    5
    5 2
    1
    3
    3
    3
    1
    2
    3
    0 0
    

    Sample Output

    CASE# 1:
    5 found at 4
    CASE# 2:
    2 not found
    3 found at 3

    题意:

           输入n和m,之后输入n个数,然后又输入m个数。输出这m个数,每个在n个数由小到大顺序中排在第几位。

    输入:

    There can be multiple test cases. Total no of test cases is less than 65. Each test case consists begins with 2 integers: N the number of marbles and Q the number of queries Mina would make. The next N lines would contain the numbers written on the N marbles. These marble numbers will not come in any particular order. Following Q lines will have Q queries. Be assured, none of the input numbers are greater than 10000 and none of them are negative.

    Input is terminated by a test case where N = 0 and Q = 0.

    输出:

    x found at y', if the first marble with number x was found at position y. Positions are numbered 1, 2,..., N.

    `x not found', if the marble with number x is not present.

    分析:

           Vector数组可以认为是大小的数组,它可以实现排序与查找功能,函数分别是sort(v.begin(),v.end() 和lower_bound(v.begin(),v.end(),x)。如果查找返回的是指针v.end()或者发现*it!=x那么容器里没有x这个数。

     1 #include <cstdio>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <vector>
     5 #include <string>
     6 #include <algorithm>
     7 #include <cmath>
     8 using namespace std;
     9 int n,m;
    10 int main(){
    11     vector<int> v;
    12     int Case = 0;
    13     while(scanf("%d%d",&n,&m) != EOF &&(n || m)){
    14         v.clear();
    15         int x;
    16         for(int i = 0 ; i < n ; i++){
    17             scanf("%d",&x);
    18             v.push_back(x);
    19         }
    20         sort(v.begin(),v.end());
    21         printf("CASE# %d:
    ",++Case);
    22         for(int i = 0 ; i < m ; i++){
    23             scanf("%d",&x);
    24             vector<int>::iterator it = lower_bound(v.begin(),v.end(),x);
    25             if(it == v.end())
    26                 printf("%d not found
    ",x);
    27             else{
    28                 if(*it == x)
    29                     printf("%d found at %d
    ",x,it - v.begin() + 1);
    30                 else
    31                     printf("%d not found
    ",x);
    32             }
    33         }
    34     }
    35     return 0;
    36 }
    View Code
  • 相关阅读:
    JAVA EE社团管理升级版-项目展示(微信小程序)
    JAVA EE社团管理升级版-微信WEB管理端说明文档
    python爬虫19 | 遇到需要的登录的网站怎么办?用这3招轻松搞定!
    python爬虫20 | 小帅b教你如何使用python识别图片验证码
    python爬虫16 | 你,快去试试用多进程的方式重新去爬取豆瓣上的电影
    python爬虫17 | 听说你又被封 ip 了,你要学会伪装好自己,这次说说伪装你的头部
    python爬虫18 | 就算你被封了也能继续爬,使用IP代理池伪装你的IP地址,让IP飘一会
    python爬虫15 | 害羞,用多线程秒爬那些万恶的妹纸们,纸巾呢?
    python爬虫13 | 秒爬,这多线程爬取速度也太猛了,这次就是要让你的爬虫效率杠杠的
    python爬虫14 | 就这么说吧,如果你不懂python多线程和线程池,那就去河边摸鱼!
  • 原文地址:https://www.cnblogs.com/cyb123456/p/5797613.html
Copyright © 2011-2022 走看看