Marbles

Gym - 101908B

Using marbles as a currency didn‘t go so well in Cubic?nia. In an attempt to make it up to his friends after stealing their marbles, the Emperor decided to invite them to a game night in his palace.

Of course, the game uses marbles, since the Emperor needs to find some use for so many of them.

• (l?u,c)(l?u,c) or;
• (l,c?u)(l,c?u) or;
• (l?u,c?u)(l?u,c?u).

Note that more than one marble can occupy the same position on the board.

As the Emperor doesn‘t like to lose, you should help him determine which games he should attend. Also, as expected, the Emperor always take the first turn when playing. Assuming both players act optimally, you are given the initial distribution of the marbles, and should find if it is possible for the Emperor to win if he chooses to play.

Input

The first line contains an integer NN (1≤N≤10001≤N≤1000). Each of the following NN rows contains two integers lili and cici indicating on which row and column the ii-th marble is in (1≤li,ci≤1001≤li,ci≤100).

Output

Your program should print a single line containing the character Y if it is possible for the Emperor to win the game or N otherwise.

Examples

2
1 3
2 3

Y

1
1 2

N
題解：
　　首先，我們需要明白一個事實，在求SG函數時，一個點的後繼中如果有必敗態就把必敗態跳過，就是說求一個點的SG函數考慮的是他所有的非必敗態的後繼。所以對於本題來說，一個點的後繼狀態中如果有坐標軸或者對角線，因為坐標軸或者對角線對於當前的先手來說是必敗的，所以他肯定不會將點移動到坐標軸或者對角線上，所以後繼為坐標軸或者對角線直接跳過，不會參與求當前點的SG值中。 坤神說 一個數的SG值一定是它的後繼中的非必敗態，或者是這個位置不知道是否為勝或敗的狀態。nim博弈中，SG值為本身是因為假設一堆石子有5個，那麽我們一定要一口氣拿完5個才能勝，否則達到1~4的狀態都是必敗的。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define inf 999
 6 using namespace std;
 7 const int maxn=110;
 8 int sg[maxn][maxn];
 9 int get_sg(int x,int y)
10 {
11     
12     if(sg[x][y]!=0)
13         return sg[x][y];
14     int vis[1100];
15     memset(vis,0,sizeof(vis));
16     for(int i=1;i<x;i++)
17     {
18         vis[get_sg(i,y)]=1;
19     }
20     for(int i=1;i<y;i++)
21     {
22         vis[get_sg(x,i)]=1;
23     }
24     for(int i=1;i<min(x,y);i++)
25     {
26         vis[get_sg(x-i,y-i)]=1;
27     }
28     for(int i=0;i<1100;i++)
29     {
30         if(!vis[i])
31         {
32             return sg[x][y]=i;
33         }
34     }
35 }
36 int main()
37 {
38     memset(sg,0,sizeof(sg));
39     sg[0][0]=0;
40     for(int i=0;i<maxn;i++)
41         sg[0][i]=inf;
42     for(int i=0;i<maxn;i++)
43         sg[i][0]=inf;
44     for(int i=1;i<maxn;i++)
45     {
46         for(int j=1;j<maxn;j++)
47         {
48             if(i==j)
49                 sg[i][j]=inf;
50         }
51     }
52     get_sg(105,104);
53     int x,y;
54     int n;
55     int flag=0;
56     int sum=0;
57     scanf("%d",&n);
58     while(n--)
59     {
60         scanf("%d%d",&x,&y);
61         if(x==y||x==0||y==0)
62         {
63             flag=1;
64         }
65         sum^=sg[x][y];
66     }
67     if(flag)
68     {
69         puts("Y");
70     }
71     else
72     {
73         if(sum)
74             puts("Y");
75         else
76             puts("N");
77     }
78     
79 }

Marbles(博弈SG函數)