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!
STL练习，map+vector。
题意是“给出一个包含n个整数的数组，你需要回答若干次询问。每次询问两个整数k和v，输出从左到右第k个v的下标（数组下标从左到右编号为1~n）”。
很显然，这是一个二维的结构，我们用下标代表要查询的数v，a[v][k]就代表第k次出现v的下标值。首先我们会想到开一个二维数组，但是这样空间消耗太大。不如利用C++STL库提供的数据结构来存储，在这里我们可以用 map+vector，申请的变量为为map <int,vector<int> > a。它的好处是存储空间是动态改变的，不用一开始就创建很大的数组来存储。
代码：
1 #pragma warning (disable:4786)
2 #include <stdio.h>
3 #include <map>
4 #include <vector>
5 using namespace std;
6 int main()
7 {
8 int i,n,m,x,k,v;
9 while(scanf("%d%d",&n,&m)!=EOF){
10 map <int,vector<int> > a;
11 for(i=0;i<n;i++){ //预处理
12 scanf("%d",&x);
13 if(!a.count(x)) //x一次没有出现过
14 a[x] = vector<int>();
15 a[x].push_back(i+1); //从0开始push
16 }
17 //查询
18 for(i=0;i<m;i++){
19 scanf("%d%d",&k,&v);
20 if(a.size()<v && a[v].size()<k)
21 printf("0
");
22 else
23 printf("%d
",a[v][k-1]); //因为是从0开始push，所以k-1
24 }
25 }
26 return 0;
27 }
Freecode : www.cnblogs.com/yym2013