HDU
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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?
"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?
51 3 2 3 3111 1 1 1 1 5 5 5 5 5 571 1 1 1 1 1 1
求出现次数超过(n+1)/2的数
很巧妙的方法,直接进行排序,输出第(n+1)/2位置上的数即可。
因为超过(n+1)/2嘛,那肯定最中间那个是它啦,不然放不开了
#include <cstdio>#include <cstring>#include<algorithm>#include <iostream>using namespace std;int a[1000000];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+1)/2]); }}
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