How many Fibs?(poj 2413)大数斐波那契

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=259#problem/C

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 <stdio.h>
#include <string.h>
#include <stdlib.h>
int a[][];
char str[][];
int main()
{
    char m[],n[];
    int i,j,sum;
    memset(a,,sizeof(a));//大数斐波那契，主要是了解思想
    a[][]=;
    a[][]=;
    a[][]=;
    for (i=;i<=;i++)
    {
        for (j=;j<=;j++)
        {
            a[i][j]=a[i][j]+a[i-][j]+a[i-][j];
            if (a[i][j]>)
            {
                a[i][j+]=a[i][j]/;
                a[i][j]=a[i][j]%;
            }
        }
    }
    int flag=,k;
    for(int i=;i<=;i++)
    {
        flag=;
        k=;
        for(int j=;j>=;j--)
        {
            if(flag||a[i][j])
            {
                flag=;
                str[i][k]=a[i][j]+'';
                k++;
            }
        }
        str[i][k]='\0';
    }
    flag=;
    /*for(i=100;i>=0;i--)
    {
        if(flag||a[100][i])
        {
            flag=1;
            printf("%d",a[100][i]);
        }
    }*/
    /*for(int i=40;i<=50;i++)
    printf("%s\n",str[i]);*/
    int l1,l2;
    while(scanf("%s%s",n,m)!=EOF)
    {
        sum=;
         l1=strlen(n);
         l2=strlen(m);
        if(n[]==''&&m[]=='') break;
        for(int i=;i<=;i++)
        {
            if((strlen(str[i])>l1&&strlen(str[i])<l2))//如果这个数的长度在范围之（a,b)长度之间，则这个数一定属于（a,b);
            {
                 sum++;
            }
            else if(l1==l2&&strlen(str[i])==l1&&strcmp(str[i],n)>=&&strlen(str[i])==l2&&strcmp(str[i],m)<=)//如果（a,b)两个数长度一样,则比较他们在字典中的大小。
            {
                  sum++;
            }
            else if(l1!=l2&&strlen(str[i])==l1&&strcmp(str[i],n)>=)
            {
                sum++;
            }
            else if(l1!=l2&&strlen(str[i])==l2&&strcmp(str[i],m)<=)
            {

                sum++;
            }
        }
        printf("%d\n",sum);;
    }
    return ;
}