uva 11991 - Easy Problem from Rujia Liu?

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

<pre name="code" class="html">#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
int main()
{
    int n,m,i,a,q,w;
	map < int,vector<int> > st;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		st.clear();    //清空map
	    for(i =0; i< n;i++)
		{
		    scanf("%d",&a);
			if(st.count(a) == 0)        //判断map里关键字a是否出现过  出现了则给它开一个vector
			{
				st[a] = vector<int> ();
			}
			st[a].push_back(i+1);       //a出现的位置入队
		}
		for(i = 0; i < m;i++)
		{
		    scanf("%d %d",&q,&w);
			if(st.count(w) == 0||st[w].size() < q)
			{
			    cout<<"0"<<endl;
			}
			else 
			{
			    cout<<st[w][q-1]<<endl;
			}
		}
	}
    return 0;
}