1095.Ignatius and the Princess IV

Time Limit: 1000 MS Memory Limit: 32768 KB
Total Submission(s): 189 Accepted Submission(s): 77

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

其实这道题就是给你n（奇数）个数，让你输出其中数量大于等于（n+1）/2的那个数。

之前在网上看了个方法叫擂台法就是求一组数中出现次数超过一半的数字，说的挺形象的叫做“打擂台”(结合代码来看），当一个数字被读入时，看一下擂台是否为空（ans是否为0），为空就占据这个擂台（ans赋值为这个数），并且记录一下这个数字在擂台上的个数（times++），接着读入，如果为同一势力（数字相同）个数增加（times++），如果不是同一势力（数字不同）占据擂台的数字就派出一个与刚读入的同归于尽（times–），如果擂台上的数字的个数减小到了0，就清空擂台（如果times减小到0，ans也要赋值为0）表示擂台为空。

//用cstdio 头文件
int main()
{
int n;
while(scanf(“%d”,&n)!=EOF)
{
int ans=0,times=0;
for(int i=0; ;i++)//这里应该是i小于n的限制条件
{
int a;
scanf(“%d”,&a);
if(ans==0||ans==a)
{
ans=a;
times++;
}
else if(ans!=a)
{
times–;
if(times==0)
ans=0;
}
}
printf(“%d\n”,ans);
}
return 0;
}

贴网址：：http://m.blog.csdn.net/smoggyxhdz/article/details/77341441