Easy Problem from Rujia Liu?(map+vector用法)

最新推荐文章于 2019-07-03 12:00:31 发布

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本文介绍了一个利用STL中的map和vector解决数组查询问题的方法。具体问题为：给定一个数组，需要回答多个查询，每个查询包含整数k和v，要求找到v在数组中第k次出现的位置。通过合理利用STL容器，有效解决了数组过大无法直接使用二维数组的问题。

题意：输入n个数，m组询问，每组询问包含两个整数k，v，询问整数v第k次出现的位置...

STL果然神奇~本来准备二维数组搞定，写了一半发现数组太大开不了，就又想到STL的map,然后还是行不通，到网上找一下，结果又发现了vector这个神奇的东东，貌似这个昨天才准备好看看的...

Easy Problem from Rujia Liu?

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

#include <stdio.h>
#include <map>
#include <vector>
using namespace std;
int main()
{
	int n,m,i,j,k,v;
	map< int,vector<int> >a;
	while(~scanf("%d%d",&n,&m))
	{
		a.clear();
		for (i=1;i<=n;i++)
		{
			scanf("%d",&j);
			a[j].push_back(i);
		}
		for (i=1;i<=m;i++)
		{
			scanf("%d%d",&k,&v);
			if (a[v].size()<k)
				printf("0\n");
			else 
				printf("%d\n",a[v][k-1]);
		}
	} 
	return 0;
}