Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others) Total Submission(s): 13419    Accepted Submission(s): 5419

Problem Description
"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
 
Author
Ignatius.L
 
 #include <stdio.h>

 typedef struct IN
{
int date;
int num;
}IN;
IN aa[];
int s[]; int cmp(const void *a,const void *b)
{
return *(int *)a - *(int *)b;
} int cmp1(const void *a,const void *b)
{
struct IN *c = (IN *)a;
struct IN *d = (IN *)b;
if(c->num!=d->num)
return c->num - d->num;
else
return c->date - d->date;
}
int main()
{
int N;
while(scanf("%d",&N)!=EOF){
int i,j,count,elem;
memset(aa,,sizeof(aa));
for(i=;i<N;i++)
scanf("%d",&s[i]);
qsort(s,N,sizeof(s[]),cmp);
elem=s[];count=;j=;i=;
while(i<N)
{
aa[j].date = elem;
while(s[i]==elem)
{
count++;
i++;
} aa[j++].num = count;
elem = s[i];
count = ;
}
//for(i=0;i<j;i++)
//printf("%d %d\n",aa[i].date,aa[i].num);
qsort(aa,j,sizeof(aa[]),cmp1);
printf("%d\n",aa[j-].date);
}
return ;
}
 
 

hdu_1029-Ignatius and the Princess IV_201310180916的更多相关文章

  1. hdu 1026(Ignatius and the Princess I)BFS

    Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J ...

  2. hdu acm 1028 数字拆分Ignatius and the Princess III

    Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ...

  3. hdu1026.Ignatius and the Princess I(bfs + 优先队列)

    Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J ...

  4. hdu 1029 Ignatius ans the Princess IV

    Ignatius and the Princess IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K ( ...

  5. ACM: HDU 1028 Ignatius and the Princess III-DP

     HDU 1028 Ignatius and the Princess III Time Limit:1000MS     Memory Limit:32768KB     64bit IO Form ...

  6. hdu 1028 Ignatius and the Princess III(DP)

    Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ...

  7. hdu 1028 Ignatius and the Princess III 简单dp

    题目链接:hdu 1028 Ignatius and the Princess III 题意:对于给定的n,问有多少种组成方式 思路:dp[i][j],i表示要求的数,j表示组成i的最大值,最后答案是 ...

  8. HDU 1027 Ignatius and the Princess II(康托逆展开)

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ( ...

  9. hdu---------(1026)Ignatius and the Princess I(bfs+dfs)

    Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J ...

  10. HDU 1029 Ignatius and the Princess IV --- 水题

    HDU 1029 题目大意:给定数字n(n <= 999999 且n为奇数 )以及n个数,找出至少出现(n+1)/2次的数 解题思路:n个数遍历过去,可以用一个map(也可以用数组)记录每个数出 ...

随机推荐

  1. 42.extjs Combobox动态加载数据问题,mode:local 还是remote

    问题: Java代码   var fabircTypeDs = new Ext.data.Store({ proxy: new Ext.data.HttpProxy({ url: 'province. ...

  2. [Swift通天遁地]三、手势与图表-(5)创建带有标题、图例、坐标轴的柱形图表

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs. ...

  3. js将字符串中所有反斜杠\替换成正斜杠/

    区分正斜杠与反斜杠: 正斜杠:http://.http紧跟着的斜杠,离手输入最近的斜杠,shift中间斜杠.45度角斜杠.正斜杠不需要转义 反斜杠:回车与空格之间.程序表示时需要转义成\\. str. ...

  4. github fork项目更改后与原作者同步更新

    1.进入你的GitHub发起Pull request 2.选择compare across forks 3.反向操作.base fork改为自己的,head fork改为原作者的 4.点击 creat ...

  5. scrapy 框架持久化存储

    1.基于终端的持久化存储 保证爬虫文件的parse方法中有可迭代类型对象(通常为列表或字典)的返回,该返回值可以通过终端指令的形式写入指定格式的文件中进行持久化操作. # 执行输出指定格式进行存储:将 ...

  6. Android内存管理(10)MAT: 基本教程

    原文: http://help.eclipse.org/mars/index.jsp?topic=%2Forg.eclipse.mat.ui.help%2Fgettingstarted%2Fbasic ...

  7. C#用Microsoft.Office.Interop.Word进行Word转PDF的问题

    之前用Aspose.Word进行Word转PDF发现'\'这个字符会被转换成'¥'这样的错误,没办法只能换个方法了.下面是Microsoft.Office.Interop.Word转PDF的方法: p ...

  8. Android互动设计-蓝牙遥控自走车iTank

    一.让Android与外部的设备互动 iTank智能型移动平台基本款简介 iTank智能型移动平台是一台履带车,车体上方的控制板有一颗微处理器,我们可以通过它的UART或是I2C接口下达指令来控制iT ...

  9. html5——web存储

    基本概念 1.传统方式我们以document.cookie来进行存储的,但是由于其存储大小只有4k左右,并且解析也相当的复杂,给开发带来诸多不便 2.h5存储设置.读取方便,而且容量较大,sessio ...

  10. JS——隐式全局变量

    在函数中,var声明的是局部变量,不带var的是隐式全局变量 <script> function fn() { var a = b = c = 0;//a是局部变量,b.c是全局变量 va ...