"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?
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
3 5 1
这道题我也不知道是不是动态规划 ,但还是在这次专题出的,所以就当做动态规划吧,题目说 这个数至少出现(N+1)/2次 那么 意思就是这个数在这奇数个数里面绝对是占了大多数 那部分 于是 我们可以吧这N个数用sort快排,给排个序,那么,最中间的那个数绝对是要求的那个数,你可以多写几个数列,你就会发现这个规律,既然知道这样,代码就很简单了,如下(注意是0~N-1还是1~N):
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1000005];
int main()
{
int N;
while(scanf("%d",&N)!=EOF)
{for(int i=0;i<N;i++)
{
scanf("%d",&a[i]);
}
sort(a,a+N);
printf("%d\n",a[N/2]);
}
return 0;
}