How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5158 Accepted Submission(s): 2007

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

思路很简单，但做得时候思维不严谨小的地方错了好多，一直在debug

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define len 10
int _cmp(char *a,char *b)    
{
    int i=0,j,count=0;
    if(strlen(a)<strlen(b))
        return 1;
    else if(strlen(a)>strlen(b))
        return -1;
    else
    {
        char x[110];
        int j=0,len1=strlen(a);
        for(i=len1-1;i>=0;i--)    /*倒置，全部都是高位在右，从右往左比较*/
            x[j++]=a[i];
        x[j]='';
        i=len1-1;
        while(i>=0)
        {
            if(b[i]>x[i])
                return 1;
            else if(b[i]==x[i])
                i--;
            else
            {
                return -1;
            }
        }
        return 0;
    }
}
char f[600][110];
int main()
{
    memset(f,'0',sizeof(f));
    int c,i,j,count,n;
    char a[101],b[101];
    f[1][0]='1';
    f[2][0]='2';
    for(i=3;i<600;i++)
    {
        c=0;
        for(j=0;j<110;j++)
        {
            f[i][j]=(f[i-2][j]+f[i-1][j]+c-2*'0')%len+'0';
            c=(f[i-2][j]+f[i-1][j]+c-2*'0')/len;
        }
    }
    for(i=1;i<600;i++)    /*变成字符串方便实用strlen（）*/
        for(j=109;j>=0;j--)
            if(f[i][j]!='0')
            {
                f[i][j+1]='';
                break;
            }
    while(cin>>a>>b)
    {
        if(a[0]=='0'&&b[0]=='0')
            break;
        count=0;
        j=0;
        for(i=1;i<=599;i++)
        {
            if(_cmp(a,f[i])>=0&&_cmp(b,f[i])<=0)
                count++;
            if(_cmp(b,f[i])>=0)
                break;
        }
        cout<<count<<endl;
    }
}