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

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <map>
     6 #include <vector>
     7 using namespace std;
     8 map <int, vector <int> > a;
     9 int main(void){
    10   int n, m;
    11 #ifndef ONLINE_JUDGE
    12   freopen("11991.in", "r", stdin);
    13 #endif
    14   while (~scanf("%d%d", &n, &m)){
    15     a.clear();
    16     int t;
    17     for (int i = 0; i < n; ++i){
    18       scanf("%d", &t);
    19       if (!a.count(t)) a[t] = vector<int>();
    20       a[t].push_back(i+1);
    21     }
    22     while (m--){
    23       int k, v;
    24       scanf("%d%d", &k, &v);
    25       if (!a.count(v) || k > a[v].size()) printf("0\n");
    26       else printf("%d\n", a[v][k-1]);
    27     }
    28   }
    29   return 0;
    30 }

    学习了一下map 的用法……

  • 相关阅读:
    Sql Server常见错误
    sql server复制需要有实际的服务器名称才能连接到服务器
    学习WCF必备网址
    利用Httphandler、UserControl 输出HTML片段
    SQL点滴—性能分析之执行计划
    我的WCF开发框架简化版及基于NET.TCP传输方式的实现
    数据库优化建议
    实现jQuery扩展总结
    【模板】zkw线段树
    洛谷 P1960 列队
  • 原文地址:https://www.cnblogs.com/liuxueyang/p/2963643.html
Copyright © 2011-2022 走看看