BigNumber_2

B - BigNumber_2

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1316

Description

Recall the definition of the Fibonacci numbers: 
f1 := 1 
f2 := 2 
fn := fn-1 + fn-2 (n >= 3) 

Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a, b]. 

 

Input

The input contains several test cases. Each test case consists of two non-negative integer numbers a and b. Input is terminated by a = b = 0. Otherwise, a <= b <= 10^100. The numbers a and b are given with no superfluous leading zeros. 

 

Output

For each test case output on a single line the number of Fibonacci numbers fi with a <= fi <= b. 

 

Sample Input

        
        
        
        

         
         
         
         10 100
1234567890 9876543210
0 0 
        
        
        
        

 

Sample Output

        
        
        
        

         
         
         
         5
4 
        
        
        
        

 

PS:又是一道大数相加+斐波那契数列

代码：

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

int ns[2100] = { 0 };

string add(string sa, string sb)
{
	string sa1, sb1;
	sa1 = sa;
	sb1 = sb;
	int la = sa.length();
	int lb = sb.length();
	if (la < lb)
	{
		string tmps = sa;
		sa = sb;
		sb = tmps;
		int tmpi;
		tmpi = la;
		la = lb;
		lb = tmpi;
	}
	string st = "";
	for (int j = lb; j < la; ++j)
	{
		st += '0';
	}
	st += sb;
	sb = st;
	int num = 0, k = 0;
	for (int m = la - 1; m >= 0; --m)
	{
		int sum = (sa[m] - '0') + (sb[m] - '0') + num;
		ns[k++] = sum % 10;
		num = sum / 10;
	}
	if (num > 0)
	{
		ns[k++] = num;
	}
	string str = "";
	for (int j = k - 1; j >= 0; --j)
	{
		str += (ns[j] + '0');
	}
	return str;
}

int main()
{
	string f[501];
	string a, b;
	f[1] = "1";
	f[2] = "2";
	for (int i = 3; i < 501; ++i)
	{
		f[i] = add(f[i - 1], f[i - 2]);
	}
	while (cin >> a >> b)
	{
		if (a[0] == '0' && b[0] == '0')
		{
			break;
		}
		int start = 0, end = 500;
		for (int i = 1; i < 501; ++i)
		{
			if (a.length() > f[i].length())
			{
				continue;
			}
			else if (a.length() < f[i].length())
			{
				start = i;
				break;
			}
			else
			{
				if (a <= f[i])
				{
					start = i;
					break;
				}
			}
		}
		for (int i = 500; i > 0; --i)
		{
			if (b.length() < f[i].length())
			{
				continue;
			}
			else if (b.length() > f[i].length())
			{
				end = i;
				break;
			}
			else
			{
				if (b >= f[i])
				{
					end = i;
					break;
				}
			}
		}
		cout << end - start + 1 << endl;
	}

	return 0;
}