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最新推荐文章于 2018-04-03 19:27:46 发布

原创最新推荐文章于 2018-04-03 19:27:46 发布 · 637 阅读

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本文介绍了一个由RujiaLiu提出的简单算法题目的解决方案，该题目要求在给定数组中找到特定整数的第k次出现的位置，并能够处理多个查询。文章提供了一段C++代码实现，通过使用映射存储每个数字出现的位置来高效解决这一问题。

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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 <iostream>
#include <cstdio>
#include <vector>
#include <map>
using namespace std;

map<int,vector<int> > a;

int main()
{
    freopen("1.in","r",stdin);
    int n,m,x,y;
    while (scanf("%d%d",&n,&m)==2) {
        a.clear();
        for (int i=0;i<n;i++) {
            scanf("%d",&x);
            if (!a.count(x)) a[x] =vector<int>();
            a[x].push_back(i+1);
        }
        while (m--){
            scanf("%d%d",&x,&y);
            if (!a.count(y) || a[y].size()<x) printf("0\n");
            else printf("%d\n",a[y][x-1]);
        }
    }
    return 0;
}