Description
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.
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.
For each query, print the 1-based location of the occurrence. If there is no such element, output 0 instead.
8 4 1 3 2 2 4 3 2 1 1 3 2 4 3 2 4 2
2 0 7 0
题目大意：
给一串数字，和k,v,输出第K个的v所在的位置。
解题思路：
利用map<int,<vector<int> > , 键值表示v , value表示一个容器，装的是同一个数字的不同的位置。
代码：
#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;//>>之间要有空格，报错了好多次才意识到。
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);//判断是否存在mymap[v].
if(it==mymap.end()||k>mymap[v].size()){
printf("0\n");
}
else{
printf("%d\n",mymap[v].at(k-1));//vector.at(),输出在k-1个位置上的数。
}
}
}
return 0;
}
UVA11991 Easy Problem from Rujia Liu?（第K个V的位置）,布布扣,bubuko.com
UVA11991 Easy Problem from Rujia Liu?（第K个V的位置）
原文地址：http://blog.csdn.net/hush_lei/article/details/38417803
