new_gray = coreMat_cmt(i,j,im) im_mask[i-4,j- 4] = new_gray else: im_mask[i-4,j-4] = im[i,j]
* 解题思路 dp[i][j]表示当n=i时,和为j出现的排列情况总数; 状态转移方程:dp[i][j]=dp[i-1][j-1]+dp[i-1][j-2]+dp[i-1][j-3]+dp[i-1][j- 6i for(int k=1;k<=6;k++){ //dp[i][j]=dp[i-1][j-1]+dp[i-1][j-2]+dp[i-1][j-3]+dp[i-1][j-
dp[i][j]表示用i个骰子扔出和为j的可能数,因为第i个骰子可能扔出1-6的点数,则dp[i][j]=dp[i-1][j-1]+dp[i-1][j-2]+dp[i-1][j-3]+dp[i-1][j-
1) = p[6] = 1,此时j - 1 = 1 ==> j = 2 ==> g(1) = 2 f(2, j) = f(5, j-1) = f(4, j-2) = f(3, j-3) = f(8, j- 4) = p[8] = 2,此时j - 4 = 1 ==> j = 5 ==> g(2) = 5 f(3, j) = f(8, j-1) = f(2, j-2) = f(5, j-3) = f(4, j-
状态转移方程是:dp[i][j] = dp[i-1][j-1] + dp[i-1][j-2] + dp[i-1][j-3] + dp[i-1][j-4] + dp[i-1][j-5] + dp[i-1]
if(status[i][j]) Value1[i][j]=0; if(status[i][j]==0) { k1=k2=0; for(i1=i,j1=j- k:k1; k1=k2=0; for(i1=i-2,j1=j-4;i1>=0&&j1>=0;i1-=2,j1-=4)//往左上数寻找电脑棋子数 { if(status k:k1; k1=k2=0; for(i1=i+2,j1=j-4;i1<=30&&j1>=2;i1+=2,j1-=4)//往左下数寻找电脑棋子数 { if(status if(status[i][j]) Value2[i][j]=0; if(status[i][j]==0) { k1=k2=0; for(i1=i,j1=j- k:k1; k1=k2=0; for(i1=i-2,j1=j-4;i1>=1&&j1>=2;i1-=2,j1-=4)//往左上数寻找玩家棋子数 { if(status
;} else { System.out.println("截取从右端开始的4~7位是:"+ch[j-7]+ch[j-6]+ch[j-5]+ch[j-4]); } }
; array[5]=""; for((i=5; i<${#perfs[*]}-2; i=i+4)) do array[i]=""; for((j=i-4; j>4; j=j-
;} else { System.out.println("截取从右端开始的4~7位是:"+ch[j-7]+ch[j-6]+ch[j-5]+ch[j-4]); } } 【程序33
;} else { System.out.println("截取从右端开始的4~7位是:"+ch[j-7]+ch[j-6]+ch[j-5]+ch[j-4]); } } 【程序33
} else { System.out.println("截取从右端开始的4~7位是:"+ch[j-7]+ch[j-6]+ch[j-5]+ch[j-
s}wO-(h-)j-4*-49-4T-5!-5O-5z-9M", "nei": "?~@;lNsE-'5-(I-/e-0!"