zoukankan      html  css  js  c++  java
  • HDU 1029 Ignatius and the Princess IV

    Ignatius and the Princess IV

      "OK, you are not too bad, em... But you can never pass the next test." feng5166 says. 

      "I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says. 

      "But what is the characteristic of the special integer?" Ignatius asks. 

      "The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says. 

      Can you find the special integer for Ignatius? 

    Input

      The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file. 
    Output

      For each test case, you have to output only one line which contains the special number you have found. 
    Sample Input

    5
    1 3 2 3 3
    11
    1 1 1 1 1 5 5 5 5 5 5
    7
    1 1 1 1 1 1 1

    Sample Output

    3
    5
    1

    解题思路:
      本题给出多组数据每组数据给出一个数n,下一行给出n个数,要求找出出现次数超过一半的数字。

    试想一个数的数量超过总数的一半其出现次数一定是所有的数中出现最多的那个数,我们就可以在输入时用一个ans记录当前出现次数最多的数,cnt记录ans比其他数的出现次数多多少。每当cnt等于0时便说明在当前输入已经输入完的数中ans所代表的数出现的次数不足一半,则cnt重新记录一个数,输入结束后获得的ans即为所取数。

      AC代码

      

    #include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n;
        while(scanf("%d", &n) != EOF){
            int ans, cnt = 0, temp;
            //ans记录当前出现次数最多的数,cnt记录ans比其他数的出现次数多多少
            //temp记录当前输入的数
            for(int i = 0; i < n; i++){
                scanf("%d", &temp);
                if(cnt == 0){//cnt等于0重新记录一个数
                    ans = temp;
                    cnt++;
                }else if(ans == temp){
                    cnt++;  //输入的数等于ans cnt++
                }else{
                    cnt--;
                }
            }
            printf("%d
    ", ans);    //输出答案
        }
        return 0;
    }

      

  • 相关阅读:
    Html5新标签解释及用法
    时间线效果
    NHibernate数据访问层核心代码
    C#开源资源项目
    安卓开发环境搭建之最新版(So Easy!)
    WIN8相关资源汇总
    HTML5 Shiv – 让该死的IE系列支持HTML5吧
    WCF 提高传输数据性能方法之二 (数据分割)
    支付宝集成问题(.net篇)
    WCF 提高传输数据性能方法之一 (使用压缩)
  • 原文地址:https://www.cnblogs.com/suvvm/p/10083113.html
Copyright © 2011-2022 走看看