How Many Fibs?

How Many Fibs?

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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

5

4

 

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=,j,count=;
    if(strlen(a)<strlen(b))
        return ;
    else if(strlen(a)>strlen(b))
        return -;
    else
    {
        char x[];
        int j=,len1=strlen(a);
        for(i=len1-;i>=;i--)    /*倒置，全部都是高位在右，从右往左比较*/
            x[j++]=a[i];
        x[j]='\0';
        i=len1-;
        while(i>=)
        {
            if(b[i]>x[i])
                return ;
            else if(b[i]==x[i])
                i--;
            else
            {
                return -;
            }
        }
        return ;
    }
}
char f[][];
int main()
{
    memset(f,'',sizeof(f));
    int c,i,j,count,n;
    char a[],b[];
    f[][]='';
    f[][]='';
    for(i=;i<;i++)
    {
        c=;
        for(j=;j<;j++)
        {
            f[i][j]=(f[i-][j]+f[i-][j]+c-*'')%len+'';
            c=(f[i-][j]+f[i-][j]+c-*'')/len;
        }
    }
    for(i=;i<;i++)    /*变成字符串方便实用strlen（）*/
        for(j=;j>=;j--)
            if(f[i][j]!='')
            {
                f[i][j+]='\0';
                break;
            }
    while(cin>>a>>b)
    {
        if(a[]==''&&b[]=='')
            break;
        count=;
        j=;
        for(i=;i<=;i++)
        {
            if(_cmp(a,f[i])>=&&_cmp(b,f[i])<=)
                count++;
            if(_cmp(b,f[i])>=)
                break;
        }
        cout<<count<<endl;
    }
}