/******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<map>

 #include<algorithm>

 #include<queue>

 #define ll __int64

 using namespace std;

 int  n;

 ll a[];

 ll b[];

 int main()

 {

     scanf("%d",&n);

     for(int i=;i<=n;i++)

         scanf("%I64d",&a[i]);

     b[n]=a[n];

     for(int i=n-;i>=;i--)

         b[i]=a[i]+a[i+];

     cout<<b[];

     for(int i=;i<=n;i++)

         cout<<" "<<b[i];

     cout<<endl;

     return ;

 }

Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string s with his directions for motion:

• An 'L' indicates he should move one unit left.
• An 'R' indicates he should move one unit right.
• A 'U' indicates he should move one unit up.
• A 'D' indicates he should move one unit down.

But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in s with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.

The first and only line contains the string s (1 ≤ |s| ≤ 100 000) — the instructions Memory is given.

If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.

In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.

In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change s to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

题意：给你一个串4个方向用4个字母表示 形成路径 要求要从起点最终回到起点 问你最少需要更改多少字母（某次的方向）

使得满足条件 回到起点。

题解：对于当前路径 判断起点与终点间的曼哈顿距离为dis

dis%2！=0则无论怎么更改都不能回到起点  （每更改一次方向都会使得距离变化2）

 /******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<map>

 #include<algorithm>

 #include<queue>

 #define ll __int64

 using namespace std;

 char s[];

 int main()

 {

     cin>>s;

     int len=strlen(s);

     int xx=,yy=;

     for(int i=; i<len; i++)

     {

         if(s[i]=='L')

             xx--;

         if(s[i]=='R')

             xx++;

         if(s[i]=='U')

             yy++;

         if(s[i]=='D')

             yy--;

     }

     if((abs(xx)+abs(yy))%)

         cout<<"-1"<<endl;

     else

         cout<<(abs(xx)+abs(yy))/<<endl;

     return ;

 }

6 3

4

8 5

3

22 4

6

 /******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<map>

 #include<algorithm>

 #include<queue>

 #define ll __int64

 using namespace std;

 int x,y;

 int a[];

 int main()

 {

     scanf("%d %d",&x,&y);

     a[]=y;a[]=y;a[]=y;

     int ans=;

     while(a[]!=x||a[]!=x||a[]!=x)

     {

         sort(a,a+);

         if(a[]+a[]->=x)

             a[]=x;

         else

             a[]=a[]+a[]-;

         ans++;

     }

     cout<<ans<<endl;

     return ;

 }