hdu 1316 How Many Fibs?

Problem 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

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

string add(string s1,string s2)
{
if(s1.length()<s2.length())
{
string temp=s1;
s1=s2;
s2=temp;
}
for(int i=s1.length()-1,j=s2.length ()-1;i>=0;i--,j--)
{
s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));
if(s1[i]-'0'>=10)
{
s1[i]=char((s1[i]-'0')%10+'0');
if(i)
s1[i-1]++;
else
s1='1'+s1;
}
}
return s1;
}

string f[1005],a,b;

int main()
{
int i;
f[1]='1';
f[2]='2';
for(i=3;i<1005;i++)
f[i]=add(f[i-2],f[i-1]);//打表
while(cin>>a>>b)
{
if(a=="0"&&b=="0")
break;
int left,right,cnt = 0,len_a,len_b,len;
len_a=a.length();
len_b=b.length();
for(i=1;i<1005;i++)
{
len=f[i].length();
if(len<len_a)
continue;
else if(len==len_a&&f[i]>=a)
{
left=i;
break;
}
else if(len>len_a)
{
left=i;
break;
}
}
for(i=1004;i>=1;i--)
{
len=f[i].length();
if(len>len_b)
continue;
else if(len==len_b&&f[i]<=b)
{
right=i;
break;
}
else if(len<len_b)
{
right=i;
break;
}
}
printf("%d\n",right-left+1);
}

return 0;
}