UVA - 11991 Easy Problem from Rujia Liu?

本文解析了RujiaLiu设置的一道简单题目,该题要求在数组中找到特定整数的第k次出现的位置。通过使用C++中的map和vector容器,文章提供了一种高效解决此问题的方法。
Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Problem E

Easy Problem from Rujia Liu?

Though Rujia Liu usually sets hard problems for contests (for example, regional contests like Xi'an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests like Rujia Liu's Presents 1 and 2), he occasionally sets easy problem (for example, 'the Coco-Cola Store' in UVa OJ), to encourage more people to solve his problems :D

Given an array, your task is to find the k-th occurrence (from left to right) of an integer v. To make the problem more difficult (and interesting!), you'll have to answer m such queries.

Input

There are several test cases. The first line of each test case contains two integers n, m(1<=n,m<=100,000), the number of elements in the array, and the number of queries. The next line contains n positive integers not larger than 1,000,000. Each of the following m lines contains two integer k and v (1<=k<=n, 1<=v<=1,000,000). The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each query, print the 1-based location of the occurrence. If there is no such element, output 0 instead.

Sample Input

8 4
1 3 2 2 4 3 2 1
1 3
2 4
3 2
4 2

Output for the Sample Input

2
0
7
0

Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!




分析:
刘汝佳大白书里的简单题。就是map,vector容器的使用。用来熟悉一下map容器。
ac代码:
#include <iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
map<int ,vector<int> >a;


int main()
{
   int n,m,k,v,x;
   while(scanf("%d%d",&n,&m)!=EOF)
   {
       a.clear();
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&x);
           if(!a.count(x))
           a[x]=vector<int>();//创建新的vector。
           a[x].push_back(i);
       }
       while(m--)
       {
          scanf("%d%d",&k,&v);
          if(!a.count(v)||a[v].size()<k)
          printf("0\n");
          else
          printf("%d\n",a[v][k-1]);
       }
   }
    return 0;
}
请帮我写一份c++代码,解决一下问题,要求符合输入输出样本There are n people (excluding myself) in my 30th birthday party. They sing the traditional “happy birthday” song: Happy birthday to you! Happy birthday to you! Happy birthday to Rujia! Happy birthday to you!!! Since I love music, I want to hear something more interesting, not that everyone sings together. Ah yes, I want one person to sing one word! For example, there are three people: Mom, Dad, Girlfriend, I’d like them to sing like this: Mom: Happy Dad: birthday Girlfriend: to Mom: you Dad: Happy Girlfriend: birthday Mom: to Dad: you Girlfriend: Happy Mom: birthday Dad: to Girlfriend: Rujia Mom: Happy Dad: birthday Girlfriend: to Mom: you Very nice, right? What if there are more than 16 people? That’s easy: repeat the song until everyone has sung at least once :) Please, don’t stop in the middle of the song.Input There is only one test case. The first line contains a single integer n (1 ≤ n ≤ 100). Then each of the next n lines contains a capitalized name (i.e. one upper-case letter followed by zero or more lowercase letters). Each name contains at most 100 characters and do not have whitespace characters inside. Output Output the song, formatted as above. Sample Input 3 Mom Dad Girlfriend Sample Output Mom: Happy Dad: birthday Girlfriend: to Mom: you Dad: Happy Girlfriend: birthday Mom: to Dad: you Girlfriend: Happy Mom: birthday Dad: to Girlfriend: Rujia Mom: Happy Dad: birthday Girlfriend: to Mom: you
最新发布
12-06
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