这场一边吃饭一边打，确实还是很菜的

C - Traveling

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0,0) at time 0, then for each ibetween 1 and N (inclusive), he will visit point (xi,yi) at time ti.

If AtCoDeer is at point (x,y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x−1,y), (x,y+1) and (x,y−1). Note that he cannot stay at his place. Determine whether he can carry out his plan.

Constraints

• 1 ≤ N ≤ 105
• 0 ≤ xi ≤ 105
• 0 ≤ yi ≤ 105
• 1 ≤ ti ≤ 105
• ti < ti+1 (1 ≤ i ≤ N−1)
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

Nt1 x1 y1t2 x2 y2:tN xN yN

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

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Sample Input 1

Copy

23 1 26 1 1

Sample Output 1

Copy

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

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Sample Input 2

Copy

12 100 100

Sample Output 2

Copy

No

It is impossible to be at (100,100) two seconds after being at (0,0).

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Sample Input 3

Copy

25 1 1100 1 1

Sample Output 3

Copy

No

 这个题目不难，就是给你坐标问你能不能到达，首先可以想想能不能到的问题，只要两点需要的步数<=你要走的步数，不直接到的话都是多走偶数步，所以只需要判断奇偶和大小就可以了

#include
       
        using namespace std;int main(){    int n,t=0,x=0,y=0,f=0;    cin>>n;    for(int i=0,tt,tx,ty;i
        
         >tt>>tx>>ty;        if((abs(tx-x)+abs(ty-y))%2!=(tt-t)%2||abs(tx-x)+abs(ty-y)>(tt-t))            f=1;        t=tt,x=tx,y=ty;    }    if(!f)cout<<"Yes";    else cout<<"No";    return 0;}
        
       

D - Checker

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

AtCoDeer is thinking of painting an infinite two-dimensional grid in a checked pattern of side K. Here, a checked pattern of side K is a pattern where each square is painted black or white so that each connected component of each color is a K × K square. Below is an example of a checked pattern of side 3:

AtCoDeer has N desires. The i-th desire is represented by xi, yi and ci. If ci is B, it means that he wants to paint the square (xi,yi) black; if ci is W, he wants to paint the square (xi,yi) white. At most how many desires can he satisfy at the same time?

Constraints

• 1 ≤ N ≤ 105
• 1 ≤ K ≤ 1000
• 0 ≤ xi ≤ 109
• 0 ≤ yi ≤ 109
• If i ≠ j, then (xi,yi) ≠ (xj,yj).
• ci is B or W.
• N, K, xi and yi are integers.

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Input

Input is given from Standard Input in the following format:

N Kx1 y1 c1x2 y2 c2:xN yN cN

Output

Print the maximum number of desires that can be satisfied at the same time.

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Sample Input 1

Copy

4 30 1 W1 2 W5 3 B5 4 B

Sample Output 1

Copy

4

He can satisfy all his desires by painting as shown in the example above.

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Sample Input 2

Copy

2 10000 0 B0 1 W

Sample Output 2

Copy

2

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Sample Input 3

Copy

6 21 2 B2 1 W2 2 B1 0 B0 6 W4 5 W

Sample Output 3

Copy

4

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D题也不难吧，但是自己比较笨，并没有做出来

先讲一下题意吧，就是你有一盘黑白棋，每个正方形黑白块的宽度都是k，但是这个黑白棋的起始位置还没有确定，你现在有一个序列，你需要知道其中最多几个有效

n是很大的啊，但是要注意到k很小，所以起始位置就没那么多了，我们需要做的是去枚举起始位置。

但是怎么去解决那个庞大的n呢，因为如果n不大的话我们就可以直接4*k*k*n的做法了，所以这个题目需要二维前缀和去查询，坐标先%一下，之后加加减减就可以了

#include
       
        using namespace std;int n,k,f[2005][2005],s[2005][2005],t;int S(int i,int j){    return i>=0&&j>=0?s[min(t-1,i)][min(t-1,j)]:0;}int get(int i,int j){    return S(i,j)-S(i,j-k)-S(i-k,j)+S(i-k,j-k);}int main(){    cin>>n>>k;    t=k*2;    string st;    for(int i=0,x,y; i
        
         >x>>y>>st;        if(st=="B")x+=k;        f[x%t][y%t]++;    }    for(int i=0; i