Project Euler:Problem 57 Square root convergents
2015-07-07 22:53
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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
√ 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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