标签：-- pre main ann ble each sam har mem
Ignatius and the Princess IV
链接：http://acm.hdu.edu.cn/showproblem.php?pid=1029
借鉴链接：https://blog.csdn.net/tigerisland45/article/details/52146154
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 43810 Accepted Submission(s):
19319
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
Author
Ignatius.L
这个题的题意就是：
输入n（n是奇数），然后输入n个整数，求其中至少出现(n+1)/2次的整数（至少有(n+1)/2个整数）。
思路：
n是奇数，(n+1)/2是n的一半以上，只要将n个数据排序，出现(n+1)/2次的整数必然会出现在中间位置。
JAVA代码：
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
@SuppressWarnings("resource")
Scanner inScanner = new Scanner(System.in);
while(inScanner.hasNext()) {
int n = inScanner.nextInt();
int[] x = new int[n]; //要掌握java的数组初始化的用法。
for(int i = 0;i<n;i++) {
x[i] = inScanner.nextInt();
}
Arrays.sort(x); //注意sort的用法，记住。
System.out.println(x[(n+1)/2]);
}
}
}
C++代码：（用了map（），思路有点复杂）
#include <iostream>
#include <map>
using namespace std;
int main()
{
int n;
map<int,int> a;
while(cin>>n)
{
a.clear();
int m;
int b=n;
while(b--)
{
cin>>m;
a[m]++;
}
for(map<int,int>::iterator it=a.begin();it!=a.end();it++)
{
if(it->second>=(n+1)/2)
{
cout<<it->first<<endl;
break;
}
}
}
return 0;
}
(Arrays.sort() 或 map) Ignatius and the Princess IV hdu1029
标签：-- pre main ann ble each sam har mem
原文地址：https://www.cnblogs.com/Weixu-Liu/p/9655255.html
