BNU20409 UVA11991

Time Limit: 1000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 11991
64-bit integer IO format: %lld      Java class name: Main

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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

https://www.bnuoj.com/v3/problem_show.php?pid=20409

题意给出一个包含n个整数的数组,你需要回答若干询问.每次询问两个整数k和v,输出从左到右第k个V的下标(数组下标从左到右1-n)

限制了内存好像没什么方法了 只有map和vector相套了

#include<stdio.h>#include<map>#include<vector>using namespace std;int main(){    map<int,vector<int> > a;//两个> >要写开    int i,x,n,m,k;    while(scanf("%d%d",&n,&m)!=EOF)    {        a.clear();        for(i=0; i<n; i++)        {            scanf("%d",&x);            if(!a.count(x))              a[x]=vector<int>();//注意 创建一个属于它的空容器            a[x].push_back(i+1);        }        while(m--)        {            scanf("%d%d",&k,&x);            if(a[x].size()<k)            {                printf("0\n");                continue;            }            printf("%d\n",a[x][k-1]);        }    }    return 0;}