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].
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
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：
View Code
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 5 int fib[700][105]; 6 7 void add( int *num1, int *num2, int *num3 ) 8 { 9 int i; 10 memset(num3, 0, 105 * sizeof(int)); 11 for( i = 0; i < 105; i++ ) 12 { 13 num3[i] += (num1[i] + num2[i]); 14 if( num3[i] >= 10 ) 15 { 16 num3[i+1]++; 17 num3[i] = num3[i] % 10; 18 } 19 } 20 } 21 int compare( int *num1, int *num2 ) 22 { 23 //for( i = 104; i >= 0; i-- ) 24 int i, len1, len2, temp; 25 for( i = 104; i >= 0; i-- ) 26 if( num1[i] != 0 ) 27 break; 28 len1 = i; 29 for( i = 104; i >= 0; i-- ) 30 if( num2[i] != 0 ) 31 break; 32 len2 = i; 33 if( len1 > len2 ) 34 return 1; 35 else if( len1 == len2 ) 36 { 37 temp = len1; 38 while( (num1[temp] == num2[temp])&&( temp != -1 ) ) 39 temp--; 40 if( temp == -1 ) 41 return 0; 42 else 43 return (num1[temp]- num2[temp]); 44 } 45 else 46 return -1; 47 } 48 49 int main(int argc, char *argv[]) 50 { 51 int i, len1, len2, cnt, left, right, n; 52 char s1[105], s2[105]; 53 int num1[105], num2[105]; 54 memset(fib[0], 0, 110 * sizeof(int)); 55 memset(fib[1], 0, 110 * sizeof(int)); 56 fib[0][0] = 1; 57 fib[1][0] = 1; 58 for( i = 2; i < 700; i++ ) 59 add(fib[i-2], fib[i-1], fib[i]); 60 while( scanf( "%s %s", s1, s2 ) != EOF ) 61 { 62 //printf( "%s %s\n", s1, s2 ); 63 if( (strcmp( s1, "0" )==0)&&(strcmp( s2, "0" )==0) ) 64 break; 65 len1 = strlen(s1); 66 len2 = strlen(s2); 67 //printf( "%d %d\n", len1, len2 ); 68 memset(num1, 0, 105 * sizeof(int)); 69 memset(num2, 0, 105 * sizeof(int)); 70 cnt = 0; 71 for( i = len1 - 1; i >= 0; i-- ) 72 num1[cnt++] = s1[i] - '0'; 73 /*for( i = 0; i < cnt; i++ ) 74 printf( "%d", num1[i] ); 75 printf( "\n" );*/ 76 cnt = 0; 77 for( i = len2 - 1; i >= 0; i-- ) 78 num2[cnt++] = s2[i] - '0'; 79 /*for( i = 0; i < cnt; i++ ) 80 printf( "%d", num2[i] ); 81 printf( "\n" );*/ 82 left = -1; 83 for( i = 1; i < 700; i++ ) 84 { 85 if( (compare( fib[i], num1 ) >= 0)&&( left == -1 ) ) 86 left = i; 87 if( compare( fib[i], num2 ) > 0 ) 88 { 89 right = i; 90 break; 91 } 92 } 93 //printf( "%d %d\n", left, right ); 94 printf( "%d\n", right - left ); 95 /*while( scanf( "%d", &n ) != EOF ) 96 { 97 for( i = 104; i >= 0; i-- ) 98 if( fib[n][i] != 0 ) 99 break; 100 for( ; i >= 0; i-- ) 101 printf( "%d", fib[n][i] ); 102 printf( "\n" ); 103 }*/ 104 } 105 //system("PAUSE"); 106 return 0; 107 }