点击打开题目
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32767 K (Java/Others) Total Submission(s): 26910 Accepted Submission(s): 11413
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
两种方法:
①STL - map存数的个数,在输入的时候就处理完成,然后输出就行了。
②把n个数sort一下,中间那个数就是结果。
代码如下:
第一种方法:
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
int main()
{
int n;
while (~scanf ("%d",&n))
{
map<int,int> ant;
int maxx = (n + 1) >> 1;
int ans;
int t;
while (n--)
{
scanf ("%d",&t);
ant[t]++;
if (ant[t] >= maxx)
ans = t;
}
printf ("%d\n",ans);
}
return 0;
}
第二种方法:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MAX 1000000
int num[MAX+11];
int main()
{
int n;
while (~scanf ("%d",&n))
{
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&num[i]);
sort(num+1,num+1+n);
printf ("%d\n",num[(n+1)>>1]);
}
return 0;
}