?Easy Problem from Rujia Liu? Time Limit:1000MS?????Memory Limit:0KB?????64bit IO Format:%lld & %llu

Description

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

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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!

題目大意：

給一串?dāng)?shù)字，和k,v,輸出第K個(gè)的v所在的位置。

解題思路：

利用map<int,<vector<int> > , 鍵值表示v , ?value表示一個(gè)容器，裝的是同一個(gè)數(shù)字的不同的位置。

代碼：

#include<iostream> #include<cstdio> #include<map> #include<vector>using namespace std;const int maxn=110000; int n,m,k,v,member; map <int,vector<int> > mymap;//>>之間要有空格，報(bào)錯(cuò)了好多次才意識(shí)到。int main(){while(scanf("%d%d",&n,&m)!=EOF){mymap.clear();for(int i=0;i<n;i++){scanf("%d",&member);mymap[member].push_back(i+1);//把位置存在vector里面。}while(m--){scanf("%d%d",&k,&v);map<int,vector<int> >::iterator it=mymap.find(v);//判斷是否存在mymap[v].if(it==mymap.end()||k>mymap[v].size()){printf("0\n");}else{printf("%d\n",mymap[v].at(k-1));//vector.at(),輸出在k-1個(gè)位置上的數(shù)。}}}return 0; }

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