算法|第十届蓝桥杯省赛B组题解记录算法|dfs|bfs|蓝桥杯

1.组队思路：从每个位置可以选取的人选入手，每个人选不可重复，组合求最大。
方法一：深搜dfs
#include using namespace std; int team[20][6],maxs=0,vis[20]; void dfs(int sum,int n)//sum为出场运动员的分数求和 n为目前要选择的号位 { if(n==6) { if(sum>maxs) maxs=sum; return ; } for(int i=0; i<20; i++) { if(!vis[i]) { vis[i]=1; dfs(sum+team[i][n],n+1); vis[i]=0; } } } int main() { freopen("201901.txt","r",stdin); //放在当前目录下面 for(int i=0; i<20; i++) { for(int j=0; j<6; j++) cin>>team[i][j]; } dfs(0,1); //因为下标0是运动员编号所以从1开始dfs cout<


 方法二：多重循环

 
#include
using namespace std;

int team[20][6],maxs=0,vis[20];
int main()
{
	freopen("201901.txt","r",stdin);
//放在当前目录下面 
	for(int i=0;
i<20;
i++)
	{
for(int j=0;
j<6;
j++)
cin>>team[i][j];

	}
	for(int i=0;
i<20;
i++)
	{
for(int j=0;
j<20;
j++)
{
for(int k=0;
k<20;
k++)
{
for(int l=0;
l<20;
l++)
{
for(int m=0;
m<20;
m++)
{
if(i!=j&&i!=k&&i!=l&&i!=m&&j!=k&&j!=l&&j!=m&&k!=l&&k!=m&&l!=m)//五个位置人员各不相同 
{
int s=team[i][1]+team[j][2]+team[k][3]+team[l][4]+team[m][5];

if(s>maxs)
maxs=s;

}
}
} 
} 
}
	}
	cout<


 
2.年号字符 思路：模拟26进制转换过程，个人认为栈很符合这个思想

 
#include
using namespace std;

stack ans;

int main()
{
	int n=2019;

	while(n)
	{
int t=n%26;

ans.push(t);

n/=26;

	}
	while(!ans.empty())
	{
putchar('A'+ans.top()-1);

ans.pop();

	}
	return 0;

}


 
3.数列求值 思路：模拟就好，没啥难的，注意取余就行

 解法一：省内存，小动一下脑子

 
#include
using namespace std;
int main()
{
	int a[10]={1,1,1};

	for(int i=3;
i<20190324;
i++)
	{
a[3]=(a[0]+a[1]+a[2])%10000;

a[0]=a[1]%10000;

a[1]=a[2]%10000;

a[2]=a[3];

	}
	cout<


 解法二：开销大，很丝滑

 
#include
using namespace std;

int a[20190324]={1,1,1};

int main()
{
	for(int i=3;
i<20190324;
i++)
	{
a[i]=(a[i-3]+a[i-2]+a[i-1])%10000;

	}
	cout<


 
4.数的分解 【算法|第十届蓝桥杯省赛B组题解记录】思路：挺简单的，暴力就行，也没啥坑

 方法一：多重for循环

 
#include
using namespace std;

int ans=0;

bool check(int n)
{
	while(n)
	{
if(n%10==2||n%10==4)
return false;

n/=10;

	}
	return true;

}
int main()
{
	int d=2019/3;

	for(int i=1;
i<=d;
i++)
	{
for(int j=i+1;
j<=2*d;
j++)
{
for(int k=j+1;
k<2019;
k++)
{
if(check(i)&&check(j)&&check(k)&&i+j+k==2019)
ans++;

}
}
	}
	cout<


 方法二：深搜dfs

 
#include
using namespace std;

int ans=0,a[1100],count1=0;

void dfs(int sum,int n,int j)
{
	if(n==3)
	{
if(sum==2019)
ans++;

return ;

	}
	for(int i=j;
i

 
5.迷宫 思路：广搜常用于求最优化路径，注意点：a.路径相同输出ASCII码最小的，设置搜索顺序时按照ASCII由小到大即可；b.不用记录最短路径，只用记录每次的方向，因为第一个找到的值一定是最小的。深搜时，需要记录每次的方向和步数，因为第一次走到的不一定是步数最少的，注意点：得到结果比较时，<即可，=的话后面的字符串序列ASCII值更大。

 解法一：bfs广搜

 
#include
using namespace std;

char a[30][50];
 
int d[4][2]={{1,0},{0,-1},{0,1},{-1,0}};
//d,l,r,u
char dr[4]={'D','L','R','U'};

int vis[30][50];
struct node{
	int x,y;

	string str;

	node(int xx,int yy,string ss)//构造函数 
	{
x=xx;

y=yy;

str=ss;

	}
};
int MAP[30][50]{//迷宫
{0,1,0,1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,0,1,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,1,0,1,0},
{0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,0,1},
{0,1,1,1,1,0,1,1,0,1,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0},
{0,1,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1},
{0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,1,0,0,0,0,0,0,0,0},
{1,1,0,0,1,0,0,0,1,1,0,1,0,1,0,0,0,0,1,0,1,0,1,1,0,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,0,1,1,1},
{0,0,0,1,1,0,1,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,1,0,0,0,0,0,0,0},
{1,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,0,1,1,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0},
{0,0,1,1,1,0,0,0,0,0,1,0,1,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,0,0,0,0,1,0,0,1},
{1,1,0,0,0,1,1,0,1,0,0,0,0,1,1,1,0,0,1,0,0,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0},
{0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,1,0,1},
{1,1,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,0,1,0,1,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,1,0,0,1,1,1,1,0,1,0,0,0,1,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,1},
{1,0,1,0,1,0,1,0,0,1,1,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,0,1,1,0,0,1,1,1,1,0,1,1,0,1,0,0,0,0,1,0,0,0},
{1,0,1,0,1,0,1,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,1,1,0,1,1,1,0,1,0,0,1},
{1,0,0,0,0,0,0,0,1,0,1,1,0,0,0,1,0,0,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0},
{1,0,1,0,1,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,0,1,0,1,0,0,1},
{0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,0,1,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,1,0,1,1,0,1,1,1,0,0,0,0,1,1,0,1,0,1},
{1,1,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,0,1,0},
{0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,1,0,1,0,1,1,0,1,0,0,0,0,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1},
{1,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,1,1,0,1,1,0,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,0,0,0,0,1},
{0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,0,1,0},
{0,0,0,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0},
{1,1,0,1,0,0,0,0,0,0,1,0,0,1,1,1,0,1,1,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1,0,0,1,0,1,1,0,1,1,1,0,1,0,0,0},
{0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,1,1,0,0,1,1},
{1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0},
{1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,1,0,0,0,0,0,0,0,1,0,0,1,0,1,0,1,0,0,0,1,0,1,1,1,0,1,0,0,0},
{0,0,1,1,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,1},
{1,0,0,0,0,0,0,1,1,0,0,1,1,1,0,1,0,1,1,1,0,1,0,0,0,1,0,0,0,1,1,0,1,1,1,0,1,0,1,0,1,1,0,1,1,1,1,0,0,0}
};

queue q;
bool check(int tx,int ty)//检查下标是否越界，是否被访问过，是否有障碍 
{
//	cout<=30||ty<0||ty>=50||vis[tx][ty]||MAP[tx][ty]==1)
return false;

	return true;

}
void bfs()
{
	node no(0,0,"");

	q.push(no);

	vis[0][0]=1;

	while(!q.empty())
	{
node t=q.front();

if(t.x==29&&t.y==49)//到达右下角 
{
cout<>a[i][j];

//	}
	
	bfs();

}


 解法二：深搜（目测代码没问题，但是因为深搜，数据又很大，很久没跑出来，有问题大家可以给我说，我的代码给大家一个参考）

 
#include
using namespace std;

char a[30][50];
 
int d[4][2]={{1,0},{0,-1},{0,1},{-1,0}};
//d,l,r,u
char dr[4]={'D','L','R','U'};

int vis[30][50];

int minStep=1000;

string minStr;
int MAP[30][50]{//迷宫
{0,1,0,1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,0,1,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,1,0,1,0},
{0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,0,1},
{0,1,1,1,1,0,1,1,0,1,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0},
{0,1,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1},
{0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,1,0,0,0,0,0,0,0,0},
{1,1,0,0,1,0,0,0,1,1,0,1,0,1,0,0,0,0,1,0,1,0,1,1,0,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,0,1,1,1},
{0,0,0,1,1,0,1,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,1,0,0,0,0,0,0,0},
{1,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,0,1,1,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0},
{0,0,1,1,1,0,0,0,0,0,1,0,1,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,0,0,0,0,1,0,0,1},
{1,1,0,0,0,1,1,0,1,0,0,0,0,1,1,1,0,0,1,0,0,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0},
{0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,1,0,1},
{1,1,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,0,1,0,1,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,1,0,0,1,1,1,1,0,1,0,0,0,1,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,1},
{1,0,1,0,1,0,1,0,0,1,1,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,0,1,1,0,0,1,1,1,1,0,1,1,0,1,0,0,0,0,1,0,0,0},
{1,0,1,0,1,0,1,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,1,1,0,1,1,1,0,1,0,0,1},
{1,0,0,0,0,0,0,0,1,0,1,1,0,0,0,1,0,0,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0},
{1,0,1,0,1,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,0,1,0,1,0,0,1},
{0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,0,1,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,1,0,1,1,0,1,1,1,0,0,0,0,1,1,0,1,0,1},
{1,1,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,0,1,0},
{0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,1,0,1,0,1,1,0,1,0,0,0,0,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1},
{1,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,1,1,0,1,1,0,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,0,0,0,0,1},
{0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,0,1,0},
{0,0,0,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0},
{1,1,0,1,0,0,0,0,0,0,1,0,0,1,1,1,0,1,1,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1,0,0,1,0,1,1,0,1,1,1,0,1,0,0,0},
{0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,1,1,0,0,1,1},
{1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0},
{1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,1,0,0,0,0,0,0,0,1,0,0,1,0,1,0,1,0,0,0,1,0,1,1,1,0,1,0,0,0},
{0,0,1,1,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,1},
{1,0,0,0,0,0,0,1,1,0,0,1,1,1,0,1,0,1,1,1,0,1,0,0,0,1,0,0,0,1,1,0,1,1,1,0,1,0,1,0,1,1,0,1,1,1,1,0,0,0}
};
bool check(int tx,int ty)//检查下标是否越界，是否被访问过，是否有障碍 
{
	if(tx<0||tx>=30||ty<0||ty>=50||vis[tx][ty]||MAP[tx][ty]==1)
return false;

	return true;

}
void dfs(int x,int y,int step,string str)
{
//	cout<=30||step>=minStep)
return;

	if(x==29&&y==49)
	{
if(step


 
6.特别数的和 思路：很简单，模拟即可，也没坑

 
#include 
using namespace std;
int main() {
int n,ans=0;

cin>>n;

for(int i=1;
i<=n;
i++)
{
int t=i;

int flag=0;

while(t)
{
if(t%10==0||t%10==2||t%10==1||t%10==9)
{
flag=1;

break;

}
t/=10;
 
}
if(flag)
ans+=i;

	}
	cout<


 
7.完全二叉树的权值 思路：考察二叉树的性质，只要知道二叉树的每层的节点数是 

文章图片 

，那就很好办，或者意识到每层的节点个数是1，2，4，....直到最后一层特别判断就好。

 
#include
using namespace std;

int t;

int main(){
	int N;

	int deep = 1;
 //深度 
	long long sum = 0;
 //每行的和 
	long long max_sum = INT_MIN;
 //最大的和 
	int max_deep = 1;

	cin>>N;

	for(int i = 1;
 i <= N;
 ++i){
cin>>t;

sum += t;

if(i == pow(2, deep) - 1){
if(max_sum < sum){ //注意不要取等号，因为题目要最小的深度 
max_deep = deep;

max_sum = sum;

}
sum = 0;

++deep;

continue;

}
if(i==N)//需要特判一下 如果不是满二叉树但是最后一层的值更大的话结果就不对
{
if(max_sum < sum){ 
max_deep = deep;

max_sum = sum;

}
}
	}
	cout<


 
8.等差数列 思路： 明确等差是相邻两个数的差值，但并不是最小的那个差值就是最后要求的等差值，因为另外比他大的等差值可能没它大，但是他俩最大公约数为1，意味着这样求出来的等差数列肯定是错的。进而就能想到所有相邻数的差值的最大公约数即为整个序列的公差，那么序列的长度=(max-min)/公差+1.

 
#include
using namespace std;

const int N=100010;

int A[N];

int n;
int gcd(int a,int b)//求最大公约数
{
	while(b)
	{
int t=a%b;

a=b;

b=t;

	}
	return a;

}int main(){
	cin>>n;

	for(int i=0;
i


 
9.后缀表达式 思路：

 
 
 
没有减号，直接求和。 
有减号的情况： 
 如果有负数存在的话：

 
 
如果负数的个数num不等于n+m+1的话，我们是可以通过加括号的形式，将所有的减号变为加号的。-5 -4 -3 -2 1,+++-,1-( 
如果负数的个数num等于n+m+1的话，肯定会留出一个负数来，无法通过加括号的形式变为正的，那么就只能加上这个负数，贪心去想，肯定是负数中最大的那个(绝对值最小的那个)。-5 -4 -3 -2 -1,+++-,-1-()，，，++--， 
 -1-（（-5+（-4）+（-3））-（-2）

 如果说没有负数的话，那么只能减去最小的那个数，剩下的就可以通过加括号的形式变为加了。如，1，2,3,4,5，两个+，两个-，则为5+4+3-（1-2）

 1,2,3,4,5 +++-,2+3+4+5-1

 1,2,3,4,5 +---,5+4-（1-2-3）

 
 坑点：注意数组开2e5+100, 因为N or M小于1e5，两个加起来开小了数组会越界；另外计算的时候注意负数相加的个数。

 
#include
using namespace std;

typedef long long LL;

const int MAX=2e5+100;

int N,M,num;

int A[MAX];

LL sum=0;

int main(){
	cin>>N>>M;

	for(int i=0;
i<=N+M;
i++)
	{
scanf("%d",&A[i]);

sum+=A[i];

if(A[i]<0)
num++;
//记录负数的个数 
	}
	sort(A,A+N+M+1);

	if(M==0)//全是+
	{
cout<


 
10.灵能传输 思路：前缀和+排序+贪心，详情思路请见yxc的视频，我只能说yyds

 坑点：暂存s0和sn时别忘了用LL，不然会溢出，另外注意处理f数组的顺序

 
#include
using namespace std;

typedef long long LL;

const int MAX=3e5+10;

int T;

LL a[MAX];

LL s[MAX],f[MAX];

bool st[MAX];

int main(){
	cin>>T;

	while(T--)
	{
int n;

cin>>n;

for(int i=1;
i<=n;
i++)
{
scanf("%lld",&a[i]);

s[i]=s[i-1]+a[i];
//计算前缀和 
}
	
LL s0=s[0],sn=s[n];

if(s0>sn)
swap(s0,sn);

sort(s,s+n+1);
for(int i=0;
i<=n;
i++)//确定s0排序后的位置 
{
if(s0==s[i])
{
s0=i;

break;

}
}for(int i=n;
i>=0;
i--)//确定sn排序后的位置从右往左是保证sn一定在s0右边 即使两数相等 
{
if(sn==s[i])
{
sn=i;

break;

}
}memset(st,false,sizeof st);
//用st数组标记哪些数字已经走过 
int l=0,r=n;

for(int i=s0;
i>=0;
i-=2)//从s0往左跳 
{
f[l++]=s[i];

st[i]=true;

}for(int i=sn;
i<=n;
i+=2)//从sn往右跳 
{
f[r--]=s[i];

st[i]=true;

}for(int i=0;
i<=n;
i++)
{
if(st[i]==false)
{
f[l++]=s[i];

}
}
LL ans=0;

for(int i=1;
i<=n;
i++)
{
ans=max(abs(f[i]-f[i-1]),ans);

}
cout<


 


		  	

    
    




    
    
    


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