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. E20170804-mk

    epic n. 史诗; 叙事诗; 史诗般的作品; estimate vt. 估计,估算; 评价,评论; 估量,估价; Sprint  vi. 冲刺,全速短跑; n. 全速短跑; 速度或活动的突然爆发; ...

  2. codevs1574广义斐波那契数列

    1574 广义斐波那契数列  时间限制: 1 s  空间限制: 256000 KB  题目等级 : 钻石 Diamond     题目描述 Description 广义的斐波那契数列是指形如an=p* ...

  3. CF1073C Vasya and Robot

    CF题目难度普遍偏高啊-- 一个乱搞的做法.因为代价为最大下标减去最小的下标,那么可以看做一个区间的修改.我们枚举选取的区间的右端点,不难发现满足条件的左端点必然是不降的.那么用一个指针移一下就好了 ...

  4. AngularJS过滤器filter-保留小数-渲染页面-小数点-$filter

    AngularJS      保留小数 默认是保留3位 固定的套路是 {{deom | number:4}} 意思就是保留小数点 的后四位 在渲染页面的时候 加入这儿个代码 用来精确浮点数,指定小数点 ...

  5. [Apple开发者帐户帮助]七、注册设备(1)注册一个设备

    您需要已注册的设备来创建开发或临时配置文件.要使用开发人员帐户注册设备,您需要拥有设备名称和设备ID. 注意:如果您使用自动签名,Xcode会为您注册连接的设备.Xcode Server也可以配置为注 ...

  6. [Swift通天遁地]二、表格表单-(18)快速应用多种预定义格式的表单验证

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

  7. python自动化测试学习笔记-5常用模块

    上一次学习了os模块,sys模块,json模块,random模块,string模块,time模块,hashlib模块,今天继续学习以下的常用模块: 1.datetime模块 2.pymysql模块(3 ...

  8. JSP执行原理图

  9. 在Windows2003安装配置Bitvise SSH Server后,不能使用软件内建立的用户登录!

    Google:  I can only log in with an administrator account - attempting to log in with a regular accou ...

  10. Android开发高手课笔记 - 01 崩溃优化(上):关于“崩溃”那点事

    Android 的两种崩溃 Java 崩溃就是在 Java 代码中,出现了未捕获的异常,导致程序异常退出 Native 崩溃一般都是因为在 Native 代码中访问非法地址,也可能是地址对齐出了问题, ...