*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 5 MB.

## Output

For each query, print the 1-based location of the occurrence. If there is no such element, output 0 instead.

Sample Input | Sample Output |

8 4 1 3 2 2 4 3 2 1 1 3 2 4 3 2 4 2 |
2 0 7 0 |

my solution:

#include <vector> #include <cstring> #include <iostream> using namespace std; vector<int> T[1000010]; int main() { int n, m; while(cin >> n >> m) { memset(T, 0, sizeof T); for(int i=1; i<=n; i++) { int temp; cin >> temp; T[temp].push_back(i); } while(m--) { int k, v; cin >> k >> v; k--; if (k<T[v].size()) cout << T[v][k] << endl; else cout << 0 << endl; } } }