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[];
int k,v;
int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
for(int i=;i<=;i++)
ve[i].clear();
for(int i=;i<=n;i++)
{
scanf("%d",&exm);
ve[exm].push_back(i);
}
for(int j=;j<=m;j++)
{
scanf("%d %d",&k,&v);
if(ve[v].size()<k||ve[v].size()==)
cout<<""<<endl;
else
cout<<ve[v][k-]<<endl;
}
}
return ;
}