project euler problem 11

最新推荐文章于 2024-04-05 11:34:21 发布

liaosharp

最新推荐文章于 2024-04-05 11:34:21 发布

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本文链接：https://blog.youkuaiyun.com/liaosharp/article/details/4573283

本文介绍了一种寻找20x20数字矩阵中四个相邻数的最大乘积的算法。通过横向、纵向及两个对角方向查找，实现了高效计算。文章详细展示了算法实现过程。

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data=""" 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48""" maxproduct=0 data=data.split() myArray=[[0 for a in range(20)] for b in range(20)] for i in range(20): for j in range(20): ind=20*i+j indexdata=int(data[ind]) myArray[i][j]=indexdata xmax=0 ymax=0 ldiag=0 rdiag=0 for i in range(20): for j in range(20): if j<=16: xtmp=myArray[i][j]*myArray[i][j+1]*myArray[i][j+2]*myArray[i][j+3] if xtmp > xmax: xmax=xtmp if i <=16: ytmp=myArray[i][j]*myArray[i+1][j]*myArray[i+2][j]*myArray[i+3][j] if ytmp > ymax: ymax =ytmp if i <=16 and j<=16: ltmp=myArray[i][j]*myArray[i+1][j+1]*myArray[i+2][j+2]*myArray[i+3][j+3] if ltmp > ldiag: ldiag=ltmp if i <=16 and j>=3: rtmp=myArray[i][j]*myArray[i+1][j-1]*myArray[i+2][j-2]*myArray[i+3][j-3] if rtmp > rdiag: rdiag=rtmp maxproduct=max(xmax,ymax,ldiag,rdiag) print maxproduct # #c is the matrix: # #m = 0 #product = lambda s: reduce(lambda a,b:a*b, s) #for i in range(16): # for j in range(16): # m = max(m,product(c[j:j+4])) # m = max(m,product([d[j] for d in mtx[i:i+4]])) # m = max(m, c[j]*c[i+1][j+1]*c[i+2][j+2]*c[i+3][j+3]) # if j >= 3 and i <= 17: # m = max(m, c[j]*c[i+1][j-1]*c[i+2][j-2]*c[i+3][j-3]) # #it is another algorithm ,very simple really