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 5MB.
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
Output for the Sample Input
2 0 7 0
Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!
给出一个包含n个整数的数组，你需要回答m个询问，每次询问两个整数k和v，输出从左到右第k个v的下标
解题思路：用map映射一个vector，对应的map<int>即为一个可变长的数组，读取数组的时候将对应值放入即可。
#include <iostream>#include <map>
#include <vector>
using namespace std;
map<int,vector<int> > q;
int main()
{
int n,m;
while(cin>>n>>m)
{
q.clear();
for(int i=0;i<n;i++)
{
int x;
cin>>x;
if(!q.count(x))
q[x]=vector<int>();
q[x].push_back(i+1);
}
for(int i=0;i<m;i++)
{
int k,v;
cin>>k>>v;
if(!q.count(v)||q[v].size()<k)
cout<<"0"<<endl;
else
cout<<q[v][k-1]<<endl;
}
}
return 0;
}
