(Problem 57)Square root convergents

时间:2021-08-25 23:45:03

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

(Problem 57)Square root convergents 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

题目大意:

2的平方根可以被表示为无限延伸的分数:

(Problem 57)Square root convergents 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

将其前四次迭代展开,我们得到:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

接下来三次迭代的展开是99/70, 239/169, and 577/408, 但是第八次迭代的展开, 1393/985, 是第一个分子的位数超过分母的位数的例子。

在前1000次迭代的展开中,有多少个的分子位数超过分母位数?

//(Problem 57)Square root convergents
// Completed on Wed, 12 Feb 2014, 04:45
// Language: C
//
// 版权所有(C)acutus (mail: acutus@126.com)
// 博客地址:http://www.cnblogs.com/acutus/
#include<stdio.h> int add(int des[],int n1,int src[],int n2){
int i,f;
for(i= , f = ; i < n1 || i < n2 ; i++){
des[i] += ( f + src[i] ) ;
f = des[i]/ ;
des[i] %= ;
}
if(f)
des[i++] = f ;
return i;
}
int main(){
int num = ,sum = , k;
int array[][] = {} ;
int nn = ,dn = , f = ;//nn分子长度,dn分母长度,f分子位置 array[][] = ;
array[][] = ;
while(num<){
//分子加分母放到分子位置成为下一个分母
k = add(array[f],nn,array[-f],dn);
//分子加分母放到分母位置成为下一个分子
nn = add( array[-f],dn,array[f],k ) ;
dn = k ;
f = - f ;
if(nn > dn) sum++;
num++;
}
printf("%d\n",sum);
return ;
}
Answer:
153