B - Ignatius and the Princess IV DP
2017-03-25 09:49
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"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?
InputThe 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.
OutputFor each test case, you have to output only one line which contains the special number you have found.
Sample Input
Sample Output
"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?
InputThe 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.
OutputFor 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 这题直接开数组也能过。还可以引入一个cnt,输入元素与上一个元素相同,cnt增加,否则cnt减少,当cnt为零时记录输入元素,因为所求数字至少出现(N+1)/2次,所以最后记录元素就是所求元素
#include <iostream> #include <string> #include<cstring> #include<cstdio> #include <vector> #include <algorithm> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { int tmp,cnt = 1,cur; scanf("%d",&tmp); for(int i=1;i<n;i++) { scanf("%d",&cur); if(cur==tmp) cnt++; else cnt--; if(cnt==0) { tmp = cur; cnt = 1; } } cout<<tmp<<endl; } }
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