UVA - 11991 - Easy Problem from Rujia Liu? （STL）

UVA - 11991

Easy Problem from Rujia Liu?

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem E

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

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!

第一次写map和vector吧，，哎，，总感觉自己慢半拍。。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;

map<int, vector<int> > a;//a[v]的'值'应当是一个数组，保存整数v从左到右依次出现的下标 

int main()
{
	int n, m, x, y;
	while(scanf("%d %d", &n, &m) == 2)
	{
		a.clear(); //clear()用于删除所有元素 
		for(int i = 0; i < n; i++)
		{
			scanf("%d", &x);
			if(!a.count(x)) a[x] = vector<int>(); //a.count返回x的出现次数，出现次数为0时给a[x]赋一个空的vector 
			a[x].push_back(i+1);// push_back()向容器末尾添加一个元素
		}
		while(m--)
		{
			scanf("%d %d", &x, &y);
			if(!a.count(y) || a[y].size() < x) printf("0\n"); //size()用于返回map中元素的个数 
			else printf("%d\n", a[y][x-1]);
		}
	}
	return 0;
}