题目描述

输入

3
4 2 3 1 10 5 9 7

输出

1

思路1:模拟+构建哈夫曼树, 当只剩下两个节点时,就是一个是冠军一个是亚军.

参考代码

#include<bits/stdc++.h>
#include<queue>
using namespace std;

typedef struct {

	int score;
	int loc;
}Node;
queue<Node> Q;
int n;
Node w1, w2;
int main() {
	cin >> n;
	int w = 1 << n;
	//cout<<w;
	for (int i = 0; i < w; i++) {
		Node node;
		cin >> node.score;
		node.loc = i + 1;
		Q.push(node);
	}
	while (Q.size() != 2) {
		w1 = Q.front();
		Q.pop();
		w2 = Q.front();
		Q.pop();
		w1.score > w2.score ? Q.push(w1) : Q.push(w2);
		//cout<<"----------------"<<endl;


	}
	w1 = Q.front();
	Q.pop();
	w2 = Q.front();
	Q.pop();
	if (w1.score > w2.score) {
		cout << w2.loc << endl;
	}
	else {
		cout << w1.loc << endl;
	}

	return 0;
}

思路2:可以把串分成前后两部分,前半部分的最高分和后半部分的最高分进行比较,胜利的便为冠军,失败的为亚军.

参考代码

#include<bits/stdc++.h>
using namespace std;
typedef struct {
	int score;
	int loc;//位置
} Node;
bool cmp(Node x,Node y) {
	return x.score<y.score;
}
Node arr[(1<<7)+10];
int n,num;
int main() {
	cin>>n;
	num = 1<<n;
	for(int i = 1; i <= num; i++) {
		cin>>arr[i].score;
		arr[i].loc = i;
	}	
	sort(arr+1,arr+num/2+1,cmp);//arr,arr+n  0  4
	sort(arr+num/2+1,arr+num+1,cmp);//arr 0 arr+1:1 arr+num/2
	
	if(arr[num/2].score > arr[num].score) {
		cout<<arr[num].loc<<endl;
	} else {
		cout<<arr[num/2].loc<<endl;
	}
	return 0;
}