UVA 11991 vector

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).
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

题意： 给你一个长度为n的序列 m组询问 第k个v在序列中的位置

题解：水 vector 处理

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
int n,m;
int exm;
vector<int> ve[1000005];
int k,v;
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=1000000;i++)
            ve[i].clear();
        for(int i=1;i<=n;i++)
            {
                scanf("%d",&exm);
                ve[exm].push_back(i);
            }
        for(int j=1;j<=m;j++)
        {
            scanf("%d %d",&k,&v);
            if(ve[v].size()<k||ve[v].size()==0)
                cout<<"0"<<endl;
            else
                cout<<ve[v][k-1]<<endl;
        }
    }
    return 0;
}