留下题目 以启发思维：
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). 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
Sample Output
2
0
7
0
难点：
Q：如何维护一个map？
A：The relationship between a map and a vector --> the relationship between a 2-D array and a 1-D array.
i.e. map由很多个动态数组构成
具体实现方法：
e.g.
#include <map> #include <vector> map<int, vector<int> >a; int main(){ a.clear(); //清空a里所有数据 if(a.count(x)==0) //判断a中以(int)x打头的vector是否为空 a[x]=vector<int>(); //在a中以(int)x打头create一个动态数组 a[x].push_back(n); //把a[x]看作一个动态数组，在其后添加元素 a[x].size() //把a[x]看作一个动态数组，计算其长度 printf("%d",a[x][n]); //把a看作一个二维数组 }
盲点：
1. 多组数据需clear
2. vector是0-based 然而题目要求的是1-based 所以输出时需特殊处理i.e.a[y][x-1]