How Many Fibs?

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Time limit: 1 Seconds   Memory limit: 32768K   
Total Submit: 56   Accepted Submit: 37   

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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

代码
#include <stdio.h>
#include <math.h>
int main()
{
    double a,b,n,sum;
    int i,j;
    scanf("%lf%lf",&a,&b);
    while(a!=0||b!=0)
    {
    sum=0;
    j=0;
     for(i=1;sum<=b;i++)
     {
         sum=((5+3*sqrt(5))/10)*pow(((1+sqrt(5)))/2,(double)(i-1))+((5-3*sqrt(5))/10)*pow(((1-sqrt(5)))/2,(double)(i-1));
         if(sum>=a)
             j++;
     }
     printf("%d\n",j-1);
    scanf("%lf%lf",&a,&b);
    }
   return 0;
}

[ 本帖最后由 ren1375342 于 2012-12-4 14:21 编辑 ]

输入区间内的斐波那契数的个数

能就这道题说一下具体怎么操作吗