UVA 11991 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!

给出一个包含n个整数的数组，你需要回答m个询问，每次询问两个整数k和v，输出从左到右第k个v的下标

解题思路：用map映射一个vector，对应的map<int>即为一个可变长的数组，读取数组的时候将对应值放入即可。

#include <iostream>
#include <map>
#include <vector>
using namespace std;

map<int,vector<int> > q;
int main()
{
int n,m;
while(cin>>n>>m)
{
q.clear();
for(int i=0;i<n;i++)
{
int x;
cin>>x;
if(!q.count(x))
  q[x]=vector<int>();
q[x].push_back(i+1);
}
for(int i=0;i<m;i++)
{
int k,v;
cin>>k>>v;
if(!q.count(v)||q[v].size()<k)
 cout<<"0"<<endl;
else
 cout<<q[v][k-1]<<endl;
}
}
return 0;
}