uva 11991Easy Problem from Rujia Liu?

原题：
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

中文：
给你n个数，然后给你m个查询，每个查询包含两个数k和v问你第k个v的小标是多少？

#include <bits/stdc++.h>using namespace std;unordered_map<int,vector<int>> miv;int n,m;int main(){    ios::sync_with_stdio(false);    while(cin>>n>>m)    {        miv.clear();        for(int i=1;i<=n;i++)        {            int res;            cin>>res;            miv[res].push_back(i);        }        for(int i=1;i<=m;i++)        {            int k,v;            cin>>k>>v;            if(miv.find(v)==miv.end()||miv[v].size()<k)                cout<<0<<endl;            else            {                cout<<miv[v][k-1]<<endl;            }        }    }    return 0;}

思路：
这题的名起的，不秒杀都不好意思~~
思路见下面的图

长方格里面的数是输入的所有数，每个方格里面的数下面的链是按顺序存储的下标值。