杭电 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

 

Source

University of Ulm Local Contest 2000

 

Recommend

Eddy

 

    此题与大菲波数那题有些类似，我在处理的时候，都是先用一个二维数组存储大菲波数。不同的是，本题需要对用个用数组存储的所谓大数进行比较，需要写一个大数比较的函数，通过该函数的返回值来判断两个大数的大小，以下是代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int fib[700][105];

void add( int *num1, int *num2, int *num3 )
{
     int i;
     memset(num3, 0, 105 * sizeof(int));
     for( i = 0; i < 105; i++ )
     {
          num3[i] += (num1[i] + num2[i]);
          if( num3[i] >= 10 )
          {
              num3[i+1]++;
              num3[i] = num3[i] % 10;
          } 
     }
}
int compare( int *num1, int *num2 )
{
    //for( i = 104; i >= 0; i-- )
    int i, len1, len2, temp;
    for( i = 104; i >= 0; i-- )
         if( num1[i] != 0 )
             break;
    len1 = i;
    for( i = 104; i >= 0; i-- )
         if( num2[i] != 0 )
             break;
    len2 = i;
    if( len1 > len2 )
        return 1;
    else if( len1 == len2 )
    {
         temp = len1;
         while( (num1[temp] == num2[temp])&&( temp != -1 ) )
                temp--;
         if( temp == -1 )
             return 0;
         else 
              return (num1[temp]- num2[temp]);
    }
    else
         return -1;
}

int main(int argc, char *argv[])
{
    int i, len1, len2, cnt, left, right, n;
    char s1[105], s2[105];
    int  num1[105], num2[105];
    memset(fib[0], 0, 110 * sizeof(int));
    memset(fib[1], 0, 110 * sizeof(int));
    fib[0][0] = 1;
    fib[1][0] = 1;
    for( i = 2; i < 700; i++ )
         add(fib[i-2], fib[i-1], fib[i]);
    while( scanf( "%s %s", s1, s2 ) != EOF )
    {
           //printf( "%s %s\n", s1, s2 );
           if( (strcmp( s1, "0" )==0)&&(strcmp( s2, "0" )==0) )
               break;
           len1 = strlen(s1);
           len2 = strlen(s2);
           //printf( "%d %d\n", len1, len2 );
           memset(num1, 0, 105 * sizeof(int));
           memset(num2, 0, 105 * sizeof(int));
           cnt = 0;
           for( i = len1 - 1; i >= 0; i-- )
                num1[cnt++] = s1[i] - '0';
           /*for( i = 0; i < cnt; i++ )
                printf( "%d", num1[i] );
           printf( "\n" );*/
           cnt = 0;
           for( i = len2 - 1; i >= 0; i-- )
                num2[cnt++] = s2[i] - '0';
            /*for( i = 0; i < cnt; i++ )
                printf( "%d", num2[i] );
           printf( "\n" );*/
           left = -1;
           for( i = 1; i < 700; i++ )
           {
                if( (compare( fib[i], num1 ) >= 0)&&( left == -1 ) )
                    left = i;
                if( compare( fib[i], num2 ) > 0 )
                {
                    right = i;
                    break;
                }
           }
           //printf( "%d %d\n", left, right );
           printf( "%d\n", right - left );   
           /*while( scanf( "%d", &n ) != EOF )
           {
                  for( i = 104; i >= 0; i-- )
                       if( fib[n][i] != 0 )
                           break;
                  for( ; i >= 0; i-- )
                       printf( "%d", fib[n][i] );
                  printf( "\n" );
           }*/
    }
  //system("PAUSE");    
  return 0;
}