zoukankan      html  css  js  c++  java
  • HDOJ4006 The kth great number 【串的更改和维护】

    The kth great number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
    Total Submission(s): 6121    Accepted Submission(s): 2471
    Problem Description
    Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.
     

    Input
    There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number. 
     

    Output
    The output consists of one integer representing the largest number of islands that all lie on one line. 
     

    Sample Input
    8 3 I 1 I 2 I 3 Q I 5 Q I 4 Q
     

    Sample Output
    1 2 3
    Hint
    Xiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000).
     

    Source
     
     
    解题思路: 
      我的想法就是,既然是要求第 k 大的数字,我就是设一数组总长度为k,其中最小的元素值即为第k大的数
      不需要优先队列 或者是 线段树、红黑树等一些高级数据结构就能写出来的一题。
     
     
    我的代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<math.h>
     4 #include<algorithm>
     5 using namespace std;
     6 int b[1000001];
     7 int main(){
     8     int i,j,t,n,k,num,flag_t,min,flag;
     9     char c;
    10     while(EOF != scanf("%d%d",&n,&k)){
    11         flag_t = 0;
    12         min = 999999999;
    13         while(n--){
    14             getchar();
    15             scanf("%c",&c);
    16             if(c == 'I'){
    17                 scanf("%d",&num);
    18                 if(flag_t < k){
    19                     b[flag_t] = num;
    20                     if(num < min){
    21                         min = num;
    22                         flag = flag_t;
    23                     }
    24                     flag_t++;
    25                 }
    26                 else if(num > min){
    27                     b[flag] = num;
    28                     min = 999999999;
    29                     for(i=0;i<k;i++){
    30                         if(b[i] < min){
    31                             min = b[i];
    32                             flag = i;
    33                         }
    34                     }
    35                 }
    36             }
    37             else if(c == 'Q'){
    38                 printf("%d
    ",min);
    39             }
    40         }
    41     }
    42     return 0;
    43 }
  • 相关阅读:
    脚本编辑器的写法
    图集优化
    数组与链表的区别
    第四课 vi编辑器使用
    第三课下 Linux termina命令行常用快捷键
    第三课上 Linux命令入门
    01.Volatile相关知识
    第二课 Ubuntu环境搭建和图形界面操作
    第一课 不要用老方法学习单片机和ARM
    第八讲 IPC之信号量Semaphore
  • 原文地址:https://www.cnblogs.com/wushuaiyi/p/3638682.html
Copyright © 2011-2022 走看看