Problem 11

# Problem_11.py
"""
In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
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
The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
四个毗邻的、同一方向的(上、下、左、右、斜)数字的最大乘积是什么?
""" def multi(li):
tot = 1
for item in li:
tot *= item
return tot raw = '''
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
'''
# 处理数据
raw = raw.split()
matrix = []
for i in range(20):
line = raw[20*i:20+20*i]
for index in range(20):
line[index] = int(line[index])
matrix.append(line) greatest_adjacent = []
greatest_product = 0
size = 4 for row in range(20):
for col in range(20):
right_product, down_product, right_dia_product, left_dia_product = 0, 0, 0, 0
right_adjacent, down_adjacent, right_dia_adjacent, left_dia_adjacent = [], [], [], []
if col <= 16:
right_adjacent = matrix[row][col:col+size]
right_product = multi(right_adjacent)
if row <= 16:
down_adjacent = [i[col] for i in matrix[row:row+size]]
down_product = multi(down_adjacent)
if col <= 16 and row <= 16:
for i in range(size):
right_dia_adjacent.append(matrix[row+i][col+i])
right_dia_product = multi(right_dia_adjacent)
if col >= 3 and row <= 16:
for i in range(size):
left_dia_adjacent.append(matrix[row+i][col-i])
left_dia_product = multi(left_dia_adjacent)
result = dict(zip([right_product, down_product, right_dia_product, left_dia_product], [right_adjacent, down_adjacent, right_dia_adjacent, left_dia_adjacent]))
compare_product = max(result)
if greatest_product < compare_product:
greatest_product = compare_product
greatest_adjacent = result[compare_product] print(greatest_adjacent, greatest_product)

Problem 11的更多相关文章

  1. leetcode problem 11 Container With Most Water

    Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). ...

  2. uoj problem 11 ydc的大树

    题目大意: 给定一颗黑白树.允许删除一个白点.最大化删点后无法与删点前距自己最远的黑点连通的黑点个数.并求出方案数. 题解: 这道题很棒棒啊. 一开始想了一个做法,要用LCT去搞,特别麻烦而且还是\( ...

  3. 状压dp找寻环的个数 Codeforces Beta Round #11 D

    http://codeforces.com/problemset/problem/11/D 题目大意:给你n个点,m条边,找该图中有几个换 思路:定义dp[i][j]表示i是圈的集合,j表示该集合的终 ...

  4. R语言学习——欧拉计划(11)Largest product in a grid

    Problem 11 In the 20×20 grid below, four numbers along a diagonal line have been marked in red. 08 0 ...

  5. Common Bugs in C Programming

    There are some Common Bugs in C Programming. Most of the contents are directly from or modified from ...

  6. electrica writeup

    关于 caesum.com 网上上的题目,分类有Sokoban,Ciphers,Maths,Executables,Programming,Steganography,Misc.题目有点难度,在努力奋 ...

  7. Kerberos简介及常见问题

    基本描述 Kerberos使用Needha-Schroeder协议作为它的基础.它使用了一个由两个独立的逻辑部分:认证服务器和票据授权服务器组成的"可信赖的第三方",术语称为密钥分 ...

  8. 【Python】Coding the Matrix:Week 5: Dimension Homework 5

    这一周的作业,刚压线写完.Problem3 没有写,不想证明了.从Problem 9 开始一直到最后难度都挺大的,我是在论坛上看过了别人的讨论才写出来的,挣扎了很久. Problem 9在给定的基上分 ...

  9. 喵哈哈村的魔法考试 Round #1 (Div.2) 题解&源码(A.水+暴力,B.dp+栈)

    A.喵哈哈村的魔法石 发布时间: 2017年2月21日 20:05   最后更新: 2017年2月21日 20:06   时间限制: 1000ms   内存限制: 128M 描述 传说喵哈哈村有三种神 ...

随机推荐

  1. servlet理解

    可得到一个结论:该JSP页面中的每个字符都由test1_jsp.java文件的输出流生成. 根据上面的JSP页面工作原理图,可以得到如下四个结论: — JSP文件必须在JSP服务器内运行. — JSP ...

  2. OSX: 安装打印机的有用命令行

    事实上非常easy.就是有用lpadmin命令,以下给出一个样例: printername="YOUR_PRINTER_NAME" location="LOCATION ...

  3. BNU 34986 Football on Table

    "Bored? Let's play table football!" The table football is played on a rectangular table, u ...

  4. UVA - 10029 Edit Step Ladders (二分+hash)

    Description Problem C: Edit Step Ladders An edit step is a transformation from one word x to another ...

  5. C语言中为什么要使用enum

    转载请注明出处,否则将追究法律责任http://blog.csdn.net/xingjiarong/article/details/47275971 在C语言中有一个关键字是enum,枚举类型,不知道 ...

  6. luogu3690 【模板】 Link Cut Tree(动态树)

    题目大意 给定n个点以及每个点的权值,要你处理接下来的m个操作.操作有4种.操作从0到3编号.点从1到n编号.0.询问从x到y的路径上的点的权值的xor和.保证x到y是联通的.1.代表连接x到y,若x ...

  7. pom.xml出现web.xml is missing and <failOnMissingWebXml> is set to true解决方案

    提示信息应该能看懂.也就是缺少了web.xml文件,<failOnMissingWebXml>被设置成true了. 搜索了一下,Stack Overflow上的答案解决了问题,分享一下. ...

  8. ubuntu16.04安装chrome谷歌浏览器

    按下 Ctrl + Alt + t 键盘组合键,启动终端. 输入以下命令: sudo wget http://www.linuxidc.com/files/repo/google-chrome.lis ...

  9. 0x55 环形与后效性问题

    poj2228 分第一天是否熟睡DP两次 #include<cstdio> #include<iostream> #include<cstring> #includ ...

  10. Uva 11021(概率)

    题意:有k只麻球,每只只能活一天,但临死之前可能产生新麻球,生出i个麻球的概率为pi,给定m,求m天后所有麻球都死亡的概率 输入格式 输入一行为测试数据的组数T,每组数据第一行为3个整数n,k,m;已 ...