hdu 1029 Ignatius and the Princess IV
Ignatius and the Princess IV
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 34652 Accepted Submission(s):
15125
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?
"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
这个题目可以很容易的看出来,在一个序列中如果去掉2个不同的元素,
那么序列中的多元素还是多元素,
所以我们只要加一个num计数器 和一个flag来保存那个元素
我们遇到一个与flag相同的num就加1 不同就减1
如果num等于0了 那就更新flag
这个题目要注意:
不能使用数组保存 不然会超内存
不能使用cin cout 不然会超时间
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdlib.h> 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <vector> 6 #include <algorithm> 7 using namespace std; 8 int main() { 9 int n; 10 while(~scanf("%d",&n)) 11 { 12 int num=0,flag,x; 13 for(int i=0;i<n;i++) 14 { 15 scanf("%d",&x); 16 if(num==0) 17 { 18 num++; 19 flag=x; 20 } 21 else if(flag==x) 22 { 23 num++; 24 } 25 else 26 { 27 num--; 28 } 29 } 30 printf("%d\n",flag); 31 } 32 return 0; 33 }
作者:红雨
出处:https://www.cnblogs.com/52why
微信公众号: 红雨python