题目大意：

求第k个v出现的索引

解题思路：

如果能构造出一个数据结构，使得data[v][k]就是第k个v出现的索引值即可求解。data[v]表示数v出现的索引数组,

data[v][k]表示第k个v出现的索引。

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

/*
 * UVA_11991.cpp
 *
 *  Created on: 2014年12月28日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <map>
#include <vector>

using namespace std;

map<int ,vector<int> > mp;

int n,m;

int main(){
	while(scanf("%d%d",&n,&m)!=EOF){
		mp.clear();

		int i;
		for(i = 0 ; i < n ; ++i){
			int tmp;
			scanf("%d",&tmp);

			/**
			 * 返回map中键k的出现次数(对于map而言，由于一个key对应一个value，
			 * 因此返回只有0和1，因此可以用此函数判断k是否在map中)
			 */
			if(mp.count(tmp) == 0){//如果这个key值没有出现过
				mp[tmp] = vector<int> ();//那么创建一个新的vector。在这里要注意vector()的实例化
			}

			mp[tmp].push_back(i+1);
		}

		while(m--){
			int k,v;
			scanf("%d%d",&k,&v);

			if(mp.count(v) == 0 || mp[v].size() < k){//如果这个值不存在,或者是需要访问的位置超过了这个只出现的次数
				printf("0\n");//打印1
			}else{
				printf("%d\n",mp[v][k-1]);
			}
		}
	}

	return 0;
}