题目57：考察2的平方根的展开

完全正确
153
[Finished in 0.1s]

fenmu, fenzi, count = [2], [3], 0
for i in range(1,1000):
        fenmu.append (fenzi[i-1]+fenmu[i-1])
        fenzi.append (fenmu[i]+fenmu[i-1])
for j in range(1000):
        if len(str(fenzi[j])) > len(str(fenmu[j])):
                count += 1
print (count)

from fractions import Fraction
count = 0
for i in range(1,1001):
    temp = 2+Fraction(1,2)
    while i:
        if i ==1:
            temp = 1+1/temp
        else:
            temp = 2+1/temp
        i -= 1
    (a,b) = str(temp).split('/')
    if len(a) > len(b):
        count += 1
print(count)

# encoding:utf-8
# 考察2的平方根展开
from time import time
def euler057(N=1000000):
    a, b = 1, 2
    count = 0
    for i in range(1, 1000):
        a, b = b, a + 2 * b
        if len(str(a + b)) > len(str(b)):
            # print(a+b,b,(a+b)/b)
            count += 1
    print(count)
if __name__ == '__main__':
    start = time() 
    euler057()
    print('cost %.6f sec' % (time() - start))

from time import time
ttt = time()
N, count, n, m = 1, 0, 1, 2
while N <= 1000:
    n, m = m, m*2 + n
    if len(str(n+m)) > len(str(m)):
        count += 1
    N += 1
print(count, time()-ttt)

%% Problem57.m
% 最后编辑时间：17-06-17 0:02
% 问题：考察2的平方根的展开
%事实上可以先求出小于等于100w的数，然后用总数减去之
% Problem57所用时间为： 3.1446秒
% Problem57的答案为： 153

function Output = Problem57()
tic

Total = 0;
Demo = 2; %分母
Elem = 3; %分子

for ii = 1:999
    NewEl = Vector_Plus(Vector_Plus(Demo,Demo), Elem);
    NewDe = Vector_Plus(Demo, Elem);
    if length(NewEl) >  length(NewDe)
        Total = Total + 1;
    end
    disp(Demo)
    Demo = NewDe;
    Elem = NewEl;
end
    
Output = Total;   
                     
toc
disp('此代码使用matlab编程')
disp(['Problem57所用时间为： ',num2str(toc),'秒'])
disp(['Problem57的答案为： ',num2str(Output)])
end
%此程序实现两个向量的加法
function Output=Vector_Plus(a,b)
if nargin==0
    a=[9 9 9 9 9 9 ];
    b=[2 0];
end
L1=length(a);
L2=length(b);
L=max(L1,L2);
if L1<L2
   Span=L2-L1;
   Newa=[zeros(1,Span),a];
   Newb=b;
elseif L1>L2
   Span=L1-L2;
   Newa=a; 
   Newb=[zeros(1,Span),b];
else
    Newa=a;
    Newb=b;
end
Rank=Newa+Newb;
for ii=L:-1:2
    if Rank(ii)>=10
        Rank(ii)=Rank(ii)-10;
        Rank(ii-1)=Rank(ii-1)+1;
    end
end
Biggest=0;
while (Rank(1)>=10)
    if Rank(1)>=10
       Rank(1)=Rank(1)-10;
       Biggest=Biggest+1;
    end
end
if Biggest>0
    Output=[Biggest,Rank];
else
    Output=Rank;
end
end
        
   

import time

def Fib(N, a, b):
    n, a, b = 0, a, b
    while n < N:
        yield a
        a, b = b,  2*b + a
        n += 1


def cal_result():
    count = 0
    fenzi = [_ for _ in Fib(1000, 3, 7)]
    fenmu = [_ for _ in Fib(1000, 2, 5)]
    for i in range(1000):
        if len(str(fenzi[i])) > len(str(fenmu[i])):
            count += 1
    return count
print("计数：{} 次\n用时：{} 秒".format(cal_result(), time.process_time()))

nume = 1
deno = 2
count = 0

for _ in range(1, 1000):
    nume, deno = deno, nume + 2 * deno
    count += len(str(nume + deno)) > len(str(deno))

print(count)

#include<iostream>
#include<vector>
using namespace std;

typedef struct BigInt{
  vector<int> v;

  BigInt(vector<int> v):v(v){}
}BI;

BI operator + (const BI & a,const BI & x){
  vector<int> t1 = a.v;
  vector<int> t2 = x.v;
  if (a.v.size() < x.v.size())  swap(t1,t2);

  for (int i = 0;i < max(t1.size(),t2.size() );i++){
      if (i < t2.size())  t1[i] += t2[i];
      if (i > 0 && t1[i-1] > 9)  {t1[i]++;  t1[i-1] -= 10;}
  }

  if (t1[t2.size()-1] > 9)
    if (t1.size() == t2.size() )  {t1.push_back(1);  t1[t1.size()-2] -= 10;}

    else {
      t1[t2.size()]++;
      for (int i = t2.size();i < t1.size()-1;i++){
        if (t1[i] > 9) {t1[i] -= 10;  t1[i+1]++;}
        else break;
      }
      t1[t2.size()-1] -= 10;
    }
  return BI(t1);
}

int main(){
  int cnt = 0;
  vector<int> a(1,1),b(2,1);
  BI n(a),d(b);

  for (int i = 0;i < 1000;i++){
    BI t = n;
    n = d;
    d = t + d + d;

    BI u = n + d;
    if (u.v.size() > d.v.size())  cnt++;
  }
  cout << cnt << endl;
  return 0;
}

from time import time

t = time()
a = 3  #a是分子
b = 2  #b是分母
l1 = []
count = 1000
while count:
    c = b #上一个数的分母
    b = a+b
    a = b+c
    if len(str(a))>len(str(b)):
        l1.append((a,b))
    count -= 1
    
print(l1,'\n')
print('用时%.6f sec,满足的数字有%d 个.' % (time()-t,len(l1)))

/*
答案：153
*/
#include <iostream>
#include <string>
#include <mpirxx.h>
using namespace std;

int main(void)
{
  mpz_class p0 = 1, p1 = 3, p2;
  mpz_class q0 = 1, q1 = 2, q2;
  int nCount = 0;
  for (int i = 2; i <= 1000; ++i)
  {
    p2 = 2 * p1 + p0;
    q2 = 2 * q1 + q0;
    string sp = p2.get_str();
    string sq = q2.get_str();
    if (sp.length() > sq.length())
      ++nCount;
    p0 = p1;
    p1 = p2;
    q0 = q1;
    q1 = q2;
  }
  cout << nCount << endl;
  return 0;
}

use std::time::Instant;

fn main() {
    let now = Instant::now();
    // Copy from "[飘飞的白杨](https://fishc.com.cn/space-uid-359393.html)"
    let (mut num, mut n, mut m) = (0, 1, 2);
    for _ in 0..1001 {
        (n, m) = (m, m * 2 + n);
        if (n + m).to_string().len() > m.to_string().len() {
            num += 1;
        }
    }
    println!("{}", num);
    println!("[Finished in {}s]", now.elapsed().as_secs_f32())
}