kuangbin专题十二 HDU1029 Ignatius and the Princess IV （水题）

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 43629    Accepted Submission(s): 19213

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

 

 

 

被专题的第一题吓到了，以为还是一道难题，就复杂考虑了。排序，然后

dp[i] = (a[i] == a[i-1]) ? dp[i-1] + 1 : 1

后来才发现必定有解，直接输出就可以了

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1000010;
int a[maxn];
int dp[maxn];

int main() {
    freopen("in.txt", "r", stdin);
    int n;
    while(~scanf("%d", &n)) { 
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
        if(n == 1) {//特判了一下
            printf("%d
", a[0]);
            continue;
        }
        sort(a, a+n);
        dp[0] = 1;
        int maxdp = dp[0], pos = -1;
        for(int i = 1; i < n; i++) {
            dp[i] = (a[i] == a[i-1]) ? dp[i-1] + 1 : 1;
            if(dp[i] > maxdp) {
                maxdp = dp[i];//最大值的dp，对应出现最多的数
                pos = i;
            }
        }
        printf("%d
", a[pos]);
    }
}