算法题：斐波那契再搞事情（Python解法）

题目

解题

这是一道填空题
难点有两个：计算和化简
巧的是
Python有可以化简分数的库函数Fraction

		>>> Fraction(10, -8)
        Fraction(-5, 4)
        >>> Fraction(Fraction(1, 7), 5)
        Fraction(1, 35)
        >>> Fraction(Fraction(1, 7), Fraction(2, 3))
        Fraction(3, 14)
        >>> Fraction('314')
        Fraction(314, 1)
        >>> Fraction('-35/4')
        Fraction(-35, 4)
        >>> Fraction('3.1415') # conversion from numeric string
        Fraction(6283, 2000)
        >>> Fraction('-47e-2') # string may include a decimal exponent
        Fraction(-47, 100)
        >>> Fraction(1.47)  # direct construction from float (exact conversion)
        Fraction(6620291452234629, 4503599627370496)
        >>> Fraction(2.25)
        Fraction(9, 4)
        >>> Fraction(Decimal('1.47'))
        Fraction(147, 100)

所以问题就可以迎刃而解

from  fractions import Fraction
n = int(input())
l=[]
ls=[]
sum=0
F1, F2 = 1, 1
l.append(F1)
l.append(F2)
for i in range(3, n + 1):
    F1, F2 = F2 % 10007, (F1 + F2) % 10007
    l.append(F2)
print(l)
for i in range(n-1):
    f=Fraction(1,l[i]*l[i+1])
    sum+=f
print(sum)

求13项的值我们需要14项斐波那契数列

官方也给出了C++解法

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

typedef __int128 lll;

lll fibo[20],mother[20],son[20];

void print(lll x){
	if(x<0){
		putchar('-');
		x=-x;
	}
	if(x>9) print(x/10);
	putchar(x%10+'0');
}

int main(){
	fibo[1]=fibo[2]=1;
	for(int i=3;i<=15;i++) fibo[i]=fibo[i-1]+fibo[i-2];
	for(int i=1;i<=13;i++) mother[i]=fibo[i]*fibo[i+1];
	lll common=1;
	for(int i=1;i<=13;i++){
		common*=mother[i];
		son[i]=1;
	}
	for(int i=1;i<=13;i++){
		lll cur=common/mother[i];
		son[i]*=cur;
	}
	lll ans=0;
	for(int i=1;i<=13;i++) ans+=son[i];
	lll g=__gcd(common,ans);
	common/=g;ans/=g;
	print(ans);
	cout<<"/";
	print(common);
	return 0;