A、B两伙马贼意外地在一片沙漠中发现了一处金矿，双方都想独占金矿，但各自的实力都不足以吞下对方，经过谈判后，双方同意用一个公平的方式来处理这片金矿。处理的规则如下：他们把整个金矿分成n段，由A、B开始轮流从最左端或最右端占据一段，直到分完为止。 马贼A想提前知道他们能分到多少金子，因此请你帮忙计算他们最后各自拥有多少金子?（两伙马贼均会采取对己方有利的策略）

第一种做法：这种方法，算法复杂性大，重复的递归
#include "stdafx.h"

#include<iostream>  
#include<vector>  
#include<string>
#include<algorithm>
#include<math.h>
#include<iomanip>
#include<numeric>
using namespace std;

int getGolds(vector<int> golds);

int delGolds(vector<int> golds)
{
	if (golds.size() == 1)
	{
		golds.erase(golds.begin());
		return 0;
	}

	int leftNum = golds[0];
	vector<int> goldsLeft = golds;
	goldsLeft.erase(goldsLeft.begin());

	int rightNum = golds[golds.size() - 1];
	vector<int> goldsRight = golds;
	goldsRight.erase(goldsRight.end() - 1);

	if (leftNum+delGolds(goldsLeft) > rightNum+delGolds(goldsRight))
	{
//		cout << "B取：" << *golds.begin() << endl;
		golds.erase(golds.begin());
		
	}
	else
	{
	//	cout << "B取：" << *(golds.end() - 1) << endl;
		golds.erase(golds.end()-1);
	}
	return getGolds(golds);
}


int getGolds(vector<int> golds)
{
	if (golds.size() == 1 )
	{
		return golds[0];
	}

	int leftNum =golds[0];
	vector<int> goldsLeft=golds;
	goldsLeft.erase(goldsLeft.begin());

	int rightNum = golds[golds.size()-1];
	vector<int> goldsRight = golds;
	goldsRight.erase(goldsRight.end() - 1);

	
	
	return max(leftNum + delGolds(goldsLeft), rightNum + delGolds(goldsRight));
}


int main()
{
	int T;
	cin >> T;
	int flag = 1;
	while (T != 0)
	{
		T--;
		vector<int> golds;
		int n;
		cin >> n;
		for (int i = 0;i < n;i++)
		{
			int g;
			cin >> g;
			golds.push_back(g);
		}

		int aSum = getGolds( golds);
		int sum = accumulate(golds.begin(),golds.end(),0);
		int bSum = sum - aSum;
		
		
		cout << "Case #" << flag << ": " << aSum << " " << bSum << endl;
	
		
//		cout << "Case #" << flag << ": " << aSumleft << " " << bSumleft << endl;
		flag++;

	}


	return 0;
}

//第二种做法，还是算法复杂性大

#include "stdafx.h"

#include<iostream>  
#include<vector>  
#include<string>
#include<algorithm>
#include<math.h>
#include<iomanip>
#include<numeric>
#include<list>
#include<deque>

using namespace std;



int getGolds(vector<int> golds,int sum)
{
	if (golds.size() == 1 )
	{
		return golds[0];
	}

	int leftNum =golds[0];
	vector<int> goldsLeft=golds;
	goldsLeft.erase(goldsLeft.begin());

	int rightNum = golds[golds.size()-1];
	vector<int> goldsRight = golds;
	goldsRight.erase(goldsRight.end() - 1);

	
	
	return sum -min(getGolds(goldsLeft,sum- leftNum),  getGolds(goldsRight, sum - rightNum));
}


int main()
{
	int T;
	cin >> T;
	int flag = 1;
	while (T != 0)
	{
		T--;
		vector<int> golds;
		int n;
		int sum=0;
		cin >> n;
		for (int i = 0;i < n;i++)
		{
			int g;
			cin >> g;
			golds.push_back(g);
			sum += g;
		}

		int aSum = getGolds( golds,sum);
		
		int bSum = sum - aSum;
		
		
		cout << "Case #" << flag << ": " << aSum << " " << bSum << endl;
	
		
//		cout << "Case #" << flag << ": " << aSumleft << " " << bSumleft << endl;
		flag++;

	}


	return 0;
}