(二分)Educational Codeforces Round 53 (Rated for Div. 2) C - Vasya and Robot

思路：这题二分长度就好了，可惜我一开始看成了只能上下或左右互换，就写炸了

做法：前缀和纪录向上和向右贡献，然后二分长度，对于长度len，枚举起点，假设起点i，终点j，j-i+1=len，那去掉区间[i,j]后计算走到了(x1,y1)，然后s=abs（x1-x）+abs(y1-y)就是要修改的路程，那只要s<=len且（s-len）%2==0就可以修改完成，（s-len）%2==0的话，说明多余的字符可以来回走消耗掉

Vasya has got a robot which is situated on an infinite Cartesian plane, initially in the cell (0,0)

. Robot can perform the following four kinds of operations:

U — move from (x,y)

to (x,y+1)

;
D — move from (x,y)
to (x,y−1)
;
L — move from (x,y)
to (x−1,y)
;
R — move from (x,y)
to (x+1,y)
- .
Vasya also has got a sequence of n
operations. Vasya wants to modify this sequence so after performing it the robot will end up in (x,y)
.

Vasya wants to change the sequence so the length of changed subsegment is minimum possible. This length can be calculated as follows: maxID−minID+1
, where maxID is the maximum index of a changed operation, and minID is the minimum index of a changed operation. For example, if Vasya changes RRRRRRR to RLRRLRL, then the operations with indices 2, 5 and 7 are changed, so the length of changed subsegment is 7−2+1=6. Another example: if Vasya changes DDDD to DDRD, then the length of changed subsegment is 1
.

If there are no changes, then the length of changed subsegment is 0

. Changing an operation means replacing it with some operation (possibly the same); Vasya can't insert new operations into the sequence or remove them.

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Help Vasya! Tell him the minimum length of subsegment that he needs to change so that the robot will go from (0,0)
to (x,y)
, or tell him that it's impossible.

Input

The first line contains one integer number n (1≤n≤2⋅105)

— the number of operations.

The second line contains the sequence of operations — a string of n

characters. Each character is either U, D, L or R.

The third line contains two integers x,y (−109≤x,y≤109)

— the coordinates of the cell where the robot should end its path.

Output

Print one integer — the minimum possible length of subsegment that can be changed so the resulting sequence of operations moves the robot from (0,0)
to (x,y). If this change is impossible, print −1
.

Examples
Input
Copy
```
5
RURUU
-2 3
```
Output
Copy
```
3
```
Input
Copy
```
4
RULR
1 1
```
Output
Copy
```
0
```
Input
Copy
```
3
UUU
100 100
```
Output
Copy
```
-1
```
Note

In the first example the sequence can be changed to LULUU. So the length of the changed subsegment is 3−1+1=3

.

In the second example the given sequence already leads the robot to (x,y)
, so the length of the changed subsegment is 0
.

In the third example the robot can't end his path in the cell (x,y)
.
```
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define maxn 1005500
ll sx[maxn],sy[maxn];int n,x,y;
char a[maxn];
int pd(int mid)
{
    int x1=0,y1=0;
    for(int i=1;i+mid-1<=n;i++)
    {
        x1=sx[i-1]+sx[n]-sx[i+mid-1];
        y1=sy[i-1]+sy[n]-sy[i+mid-1];
        int s=abs(x1-x)+abs(y1-y);
        if(mid>=s&&(mid-s)%2==0) return 1;
    }
    return 0;
}
int main()
{

    scanf("%d",&n);
    scanf("%s",a+1);
    scanf("%d %d",&x,&y);
    for(int i=1;i<=n;i++)
    {
        if(a[i]=='U') sy[i]=sy[i-1]+1;
        else if(a[i]=='D') sy[i]=sy[i-1]-1;
        else sy[i]=sy[i-1];
        if(a[i]=='R') sx[i]=sx[i-1]+1;
        else if(a[i]=='L') sx[i]=sx[i-1]-1;
        else sx[i]=sx[i-1];
    }
    int l=0,r=n,p=-1;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(pd(mid))
        {
            p=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    printf("%d\n",p);
}
```

(二分)Educational Codeforces Round 53 (Rated for Div. 2) C - Vasya and Robot

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