zoukankan      html  css  js  c++  java
  • UVA-11991 Easy Problem from Rujia Liu?


    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 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
    

    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!

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <map>
     5 
     6 using namespace std;
     7 
     8 int main()
     9 {
    10     int n,m;
    11     while(scanf("%d%d",&n,&m)!=EOF)
    12     {
    13         int a,b,x;
    14         map< int , int > mII;
    15         map< pair<int ,int > , int > ans;
    16         for(int i=1;i<=n;i++)
    17         {
    18             scanf("%d",&x);
    19             mII[x]++;
    20             ans[ make_pair( x , mII[x] ) ]=i;
    21         }
    22         while(m--)
    23         {
    24             scanf("%d%d",&a,&b);
    25             printf("%d
    ",ans[make_pair(b,a)]);
    26         }
    27     }
    28     return 0;
    29 }
  • 相关阅读:
    面向对象简述
    python面向对象
    Python中int()函数的用法浅析
    给定一个字符串 s,找到 s 中最长的回文子串。你可以假设 s 的最大长度为1000。
    python中关于round函数的小坑
    Xmind8破解,以及相关的流程和破解包
    生成器和生成器表达式
    brush图标
    js声明全局变量的方式
    js修改全局变量
  • 原文地址:https://www.cnblogs.com/CKboss/p/3407279.html
Copyright © 2011-2022 走看看