题目42：这个英语词列表里共有多少个三角形词？

欧拉计划 · 发表于 2015-5-2 12:11:29

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本帖最后由欧拉计划于 2015-5-2 12:20 编辑

Coded triangle numbers

The term of the sequence of triangle numbers is given by, ; so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?

题目：

三角形数序列中第 n 项的定义是：

；此前十个三角形数是：

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

通过将一个单词中每个字母在字母表中的位置值加起来：我们可以将一个单词转换为一个数。例如，单词 SKY 的值为

如果单词的值是一个三角形数，我们称这个单词为三角形单词。

p042_words.txt (15.96 KB, 下载次数: 48) （右键另存为）是一个 16K 的文本文件，包含将近两千个常用英语单词。在这个文件中，一共有多少个三角形词？

迷雾少年 · 发表于 2016-8-30 13:58:52

不是很懂题目意思。。

愤怒的大头菇 · 发表于 2016-9-7 12:49:38

dic = {
'A':1,
'B':2,
'C':3,
'D':4,
'E':5,
'F':6,
'G':7,
'H':8,
'I':9,
'J':10,
'K':11,
'L':12,
'M':13,
'N':14,
'O':15,
'P':16,
'Q':17,
'R':18,
'S':19,
'T':20,
'U':21,
'V':22,
'W':23,
'X':24,
'Y':25,
'Z':26}
def San():
list1 = []
for i in range(10000):
list1.append(i*(i+1)/2)
return list1
list1 = []
f = open('word.txt')
for word in f:
str_1 = ''
for each in word:
if each != '"':
str_1 += each
list1.append(str_1)
count = 0
list2 = []
for each in list1:
for i in each:
if i == ',':
list2.append(count)
count = 0
else:
count += dic[i]
list2.append(count)
list3 = San()
total = 0
for each in list2:
if each in list3:
total += 1
print(total)

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答案：162

飘飞的白杨 · 发表于 2016-10-8 11:31:40

count = 0 #用于记数
num = lambda s: sum(int(ord(i))-64 for i in s) #计算每个单词的值
list0 = [0.5*i*(i+1) for i in range(100)] #生成一个三角数的list
with open("p042_words.txt","r") as f:
text = f.read()
for x in text.replace(""","").split(","): #对文本处理，生成单词列表
if num(x) in list0:
count += 1
print(count)

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芒果加黄桃 · 发表于 2017-1-15 11:43:16

# encoding:utf-8
# 查找三角形单词个数
from time import time
def euler042():
l_trinum = [int(i * (i + 1) / 2) for i in range(30)]
d_abc = {'A':1, 'B':2, 'C':3, 'D':4, 'E':5, 'F':6, 'G':7, 'H':8, 'I':9, 'J':10, 'K':11, 'L':12, 'M':13, 'N':14, 'O':15, 'P':16, 'Q':17, 'R':18, 'S':19, 'T':20, 'U':21, 'V':22, 'W':23, 'X':24, 'Y':25, 'Z':26}
count = 0
with open('p042_words.txt') as f:
l_text = f.read().replace('"', '').split(',')
for each in l_text:
w_n = 0
for t in [w for w in list(each)]:
w_n += d_abc.get(t)
if w_n in l_trinum:
count += 1
print(count)
if __name__ == '__main__':
start = time()
euler042()
print('cost %.6f sec' % (time() - start))

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162
cost 0.007988 sec

渡风 · 发表于 2017-6-12 22:44:37

%% Problem42.m
%最后编辑时间：2017-06-12 22:10 版本1.0
%找出转换值为三角形数值得单词数
% Problem42所用时间为0.17588秒
% Problem42的答案为162

%% Problem42.m
%最后编辑时间：2017-06-12 22:10 版本1.0
%找出转换值为三角形数值得单词数
% Problem42所用时间为0.17588秒
% Problem42的答案为162
function Output = Problem42()
tic
Output = 0;
Data = importdata('p042_words.txt');
Data = cell2mat(Data);
Qua = 0; %引号个数
QuaLoca = []; %引号位置
for ii = 1:length(Data)
if strcmp(Data(ii),'"') == 1
Qua = Qua + 1;
QuaLoca = [QuaLoca,ii];
end
end
Words = cell(1, Qua /2);
for jj = 1:length(Words)
Words{jj} = Data(QuaLoca(jj*2-1)+1 : QuaLoca(jj*2)-1); %提取两个引号之间的字母
if IsTriNum(Words{jj}) == 1;
Output = Output + 1;
end
end
toc
disp('此代码使用matlab编程')
disp(['Problem42所用时间为',num2str(toc),'秒'])
disp(['Problem42的答案为',num2str(Output)])
end
%% 输入一个字母判定其是否为判定它是否为三角形数
function Output = IsTriNum(Input)
if nargin == 0
Input = 'JK';
end
Letter = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
InputScore = 0;
for ii = 1:length(Input)
for jj = 1:26
if strcmp(Input(ii),Letter(jj))
InputScore = InputScore + jj;
break
end
end
end
n = floor(sqrt(InputScore * 2));
if InputScore * 2 == n*(n+1)
Output = 1;
else
Output = 0;
end
end

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najin · 发表于 2017-9-30 10:54:03

用的matlab
结果是：
>> Untitled

count =

162

时间已过 0.007360 秒。
>>

k往事如烟k · 发表于 2019-3-29 14:55:29

def t(n):
return n*(n+1)//2
path = r"E:\Python EX\Euler Project\p042_words.txt"
with open(path, "r") as f:
words = f.read()
wordsList = words.strip('"').split('","')
wordsValueList = []
for each in wordsList:
sum = 0
for i in range(len(each)):
sum += (ord(each[i])-ord("A")+1)
wordsValueList.append(sum)
triNumList = []
for i in range(1, 100):
if t(i) <= max(wordsValueList):
triNumList.append(t(i))
else:
break
finalWordsList = []
for each in wordsList:
sum = 0
for i in range(len(each)):
sum += (ord(each[i])-ord("A")+1)
if sum in triNumList:
finalWordsList.append(each)
print(len(finalWordsList))

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162.....写的感觉有点重复~~~太丑了

王小召 · 发表于 2019-6-14 15:06:52

本帖最后由永恒的蓝色梦想于 2020-7-2 18:39 编辑

共有：162 个
用时：0.1404009 秒

import math
import functools
import time
def trangle_num(num):
if int(math.sqrt(8*num + 1)) == math.sqrt(8*num + 1):
return True
def cal_result():
count = 0
with open(r"C:\Users\wangyongzhao\Desktop\p042_words.txt") as f:
for each_word in [each.split('"')[1] for each in f.read().split(",")]:
if trangle_num(functools.reduce(lambda x, y: x + y, [ord(i) - 64 for i in each_word])):
count += 1
return count
print("共有：{} 个\n用时：{} 秒".format(cal_result(), time.process_time()))

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debuggerzh · 发表于 2020-8-11 17:29:32

162

Process returned 0 (0x0) execution time : 0.054 s
Press any key to continue.
又可以利用Excel VBA处理源文件使之易于输入

Public Sub PE42()
Open "C:\Users\zcfcf\Desktop\words.txt" For Input As #1
Dim s As String
Dim cnt As Integer
cnt = 1
Do Until EOF(1)
Input #1, s
Range("a" & CStr(cnt)).Value = s
cnt = cnt + 1
Loop
Close #1
End Sub

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接着利用ASCII码天然的编码特性计算单词的值，并加以统计

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
const int a[] = {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136,
153, 171, 190, 210, 231, 253, 276, 300, 325, 351};
int calc(const string & s){
int res = 0;
for (int i = 0;i < s.length();i++)
res += s[i] - 'A' + 1;
return res;
}
int main(){
freopen("i.in","r",stdin);
string s;
int cnt = 0;
while(cin >> s){
int t = calc(s);
for (int i = 0;a[i] <= 351;i++)
if (t == a[i]) { cnt++; break; }
}
cout << cnt << endl;
return 0;
}

复制代码

永恒的蓝色梦想 · 发表于 2021-1-11 15:54:56

本帖最后由永恒的蓝色梦想于 2021-3-6 10:40 编辑

#include<iostream>
#include<fstream>
#include<cmath>
int main() {
using namespace std;
ios::sync_with_stdio(false);
ifstream file("p042_words.txt");
unsigned int val, count = 0;
char ch;
while (file) {
if (file.get() == '"') {
if (file) {
for (val = 0; (ch = file.get()) != '"'; val += ch - 'A' + 1);
count += !fmod(sqrt((val << 3) | 1), 1.0);
}
else {
break;
}
}
}
file.close();
cout << count << endl;
return 0;
}

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a1351468657 · 发表于 2021-3-20 12:17:55

#include <stdio.h>
#include <math.h>
int check_trinum(int);
int check_trinum(int num)//判断是否为三角形数
{
int i, j, k;
j = sqrt(2 * num);
for (i = j; i >= 1; i--)
{
k = i * (i + 1) / 2;
if (k == num)
{
return 1;
}
}
return 0;
}
main()
{
FILE *fp;
int i = 0, j, k, n, count = 0;
char ch, word[3000][20];
fp = fopen("words.txt", "r");
while (!feof(fp))//将文件中名字写入字符串数组word中
{
ch = fgetc(fp);
if (ch == '"')
{
j = 0;
continue;
}
else if (ch == '\n')
{
continue;
}
else
{
if (ch == ',')
{
i++;
}
else if (ch >= 'A' && ch <= 'Z')
{
word[i][j] = ch;
word[i][j + 1] = '\0';
j++;
}
}
}
fclose(fp);
k = i;
for (i = 0; i <= k; i++)
{
n = 0;
for (j = 0; word[i][j] != '\0'; j++)
{
n += (word[i][j] - 64);//因为'A'的acsii码为65，所以减去64就是字母在字母表的位置
}
if (check_trinum(n))
{
count++;//记录三角形词的个数
}
}
printf("%d", count);
}

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162

ft215378 · 发表于 2021-10-21 19:25:26

#共有多少个三角形词
'''
角形数序列中第 n 项的定义是：Tn = 1/2n(n+1)；此前十个三角形数是：
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
'''
from time import *
#生成n项三角形数序列
def creat_triangle_numlist(n):
triangle_numlist = []
for i in range(1, n+1):
      triangle_numlist.append((i * (i+1)) // 2)
return triangle_numlist

def get_letter_value(letter):
letter_dic = {'A':1, 'B':2, 'C':3, 'D':4, 'E':5, 'F':6, 'G':7, 'H':8, 'I':9\
               , 'J':10, 'K':11, 'L':12, 'M':13, 'N':14, 'O':15, 'P':16, \
               'Q':17 , 'R':18, 'S':19, 'T':20, 'U':21, 'V':22, 'W':23, \
               'X':24, 'Y':25, 'Z':26}
return letter_dic[letter]

#处理文件获得文本
with open("c:\\Users\\Administrator\\Desktop\\word.txt") as f:
word_str = f.read()

start = time()
#处理字符串
word_list = word_str.split(',')

#计算多少三角形词
count = 0
each_sum = 0
word_value = []
for each in word_list:
each = each.replace('"', '')
for i in each:
      each_sum += get_letter_value(i)
word_value.append(each_sum)
each_sum = 0
#print(max(word_value)) 129160
#'''

triangle = creat_triangle_numlist(1000)
result = []
for each in word_value:
if each in triangle:
      result.append(each)
      count += 1
else:
      continue
end = time()
print(count)
print("用时%4f秒" % (end-start))

B1tetheDust · 发表于 2022-10-24 19:55:56

import time as t
import numpy as np
start = t.perf_counter()
def is_triangle_number(tn):
n = int(np.sqrt(2 * tn))
if 2 * tn == n * (n + 1):
return True
else:
return False
names = np.loadtxt("C:/Users/wuhw/Desktop/p042_words.txt", dtype=str, delimiter=',')
name_list = []
count_nums = 0
for name in names:
name_list.append(name.strip('"'))
for each in name_list:
name_value = sum([ord(i) - 64 for i in each])
if is_triangle_number(name_value):
count_nums += 1
print(count_nums)
print("It costs %f s" % (t.perf_counter() - start))

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162
It costs 0.009455 s

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题目42：这个英语词列表里共有多少个三角形词？

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