Project Euler：Problem 57 Square root convergents

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

√ 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?

找规律，大数加法

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

string strplus(string a, string b)
{
	int lena = a.length();
	int lenb = b.length();
	int len;
	if (lena <= lenb)
	{
		int gap = lenb - lena;
		string c;
		c.resize(gap, '0');
		a = c + a;
		len = lenb;
	}
	else
	{
		int gap = lena - lenb;
		string c;
		c.resize(gap, '0');
		b = c + b;
		len = lena;
	}
	int flag = 0;
	string ans = "";
	for (int i = len - 1; i >= 0; i--)
	{
		int tmp = a[i] + b[i] - '0' - '0' + flag;
		flag = tmp / 10;
		tmp = tmp % 10;
		char p = tmp + '0';
		ans = p + ans;
	}
	if (flag == 1)
		ans = '1' + ans;
	return ans;
}

int main()
{
	string a = "3";
	string b = "2";
	int count = 0;
	for (int i = 2; i <= 1000; i++)
	{
		string tmp = b;
		b = strplus(a, b);
		a = strplus(b, tmp);
		if (a.length() > b.length())
			count++;
	}
	cout << count << endl;
	system("pause");
	return 0;
}

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