UVA 11991 STL中map、vector的应用

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3142

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 41 3 2 2 4 3 2 11 32 43 24 2

Output for the Sample Input

2070

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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!

题目大意：给出包含n个整数的数组，你需要回答若干询问，每次询问的个数是两个整数，k和v，输出从左到右第k个v的下标（数组的编号从1到n）

解题思路：见代码

#include <stdio.h>#include <vector>#include <map>#include <vector>using namespace std;map<int ,vector<int> >a;//最后两个 > 不要连写，否则会被误认为 >>int main(){    int m,n,x,y;    while(~scanf("%d%d",&n,&m))    {        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;}