正題

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題目大意

$n?n$的矩形，每一個人操作時如果棋盤上有一個$k?k$的矩形空地就可以選擇一個點堵上。如果沒有就失敗了，求必勝方。

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解題思路

如果場地上有一個位置堵上后即可堵上所有$k?k$的矩形那么這個點被堵住后就贏了，所以先手必勝。

如果沒有這個位置，那么最后場地上剩下一個$k?k$的矩形時先手必勝，也就是如果我們找出兩個不相交的$k?k$矩形那么顯然先手不會去堵上這兩個中的任何一個，也就是其他位置會被優先堵上。所以只要判斷其他格子數量的奇偶性即可。

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$code$

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=1100; int n,k,a[N][N],b[N][N]; char s[N]; bool check(int x,int y) {x--;y--;return (a[x+k][y+k]+a[x][y]-a[x][y+k]-a[x+k][y])==0;} int main() { // freopen("lcyrcx.in","r",stdin); // freopen("lcyrcx.out","w",stdout);scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%s",s+1);for(int j=1;j<=n;j++)a[i][j]=a[i-1][j]+a[i][j-1]+(s[j]=='1')-a[i-1][j-1];}int z=0;for(int i=1;i<=n-k+1;i++)for(int j=1;j<=n-k+1;j++)if(check(i,j)){b[i][j]++;b[i+k][j+k]++;b[i+k][j]--;b[i][j+k]--;z++;}bool flag=1;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];if(b[i][j]==z)flag=0;}if(flag){if((a[n][n]-2*k*k)&1)printf("rx");else printf("yc");}else{if(z)printf("rx");else printf("yc");} } /* 100 20 0000000000000000000000000000000000000000010100000100001111110011011010100110010001111000110010000101 0000000000000000000000000000000000000000111000011001101111011011100100000000110110000011001111001001 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