How many Fibs?（高精度）

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<=10¹⁰⁰. 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 1001234567890 98765432100 0

Sample Output

54

解题思路

先打个斐波那契的表，我打了480个，感觉高精度的题必须自己动手多操作，细节很多，看别人的代码只要看个思路，细看就不值得了，细节还是自己操作时，一步步调试，

下面的代码写得略拙-_-!,，本来是用二分查找的，楸了半天没楸好，过两天再试试。

特注：### strcmp()函数并不是先比较字符串长度的，而是直接相减 。

AC代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxnlen = 210;char fib[485][maxnlen];                                        //存放斐波那契数列void turn(char *a)                                             //定义倒序函数{    int L;    char temp[maxnlen];    L=strlen(a);    for(int i=L-1, j=0; i>=0; i--, j++)        temp[j]=a[i];    strcpy(a, temp);    a[L] = '\0' ;}void get_fibs(){    int pp, i, j;    strcpy(fib[1], "1");    strcpy(fib[2], "2");    for(i=3; i<485; i++)    {        pp = 0;        int L1 = strlen( fib[i-2] );        int L2 = strlen( fib[i-1] );        for(j=0; j<L1; j++)        {            fib[i][j] = (fib[i-2][j] - '0' + fib[i-1][j] - '0' + pp) % 10 + '0' ;            pp = (fib[i-2][j] - '0' + fib[i-1][j] - '0' + pp) / 10 ;        }        if( L1 < L2 )            fib[i][j] = (fib[i-1][j] - '0' + pp) + '0' ;        else if( L1==L2 && pp == 1 )            fib[i][j] = '1' ;                                   //fib[]中一直是倒序的    }    for(int i=1; i<485; i++)        turn( fib[i] );                                         //转为正序}int main(){    get_fibs();    int i, j, k, flag1, flag2, flag=0;    char a[maxnlen], b[maxnlen];    while( scanf("%s%s", a, b) )                                           //有待优化的查找！！！    {        if( strcmp(a, "0")==0 && strcmp(b, "0")==0 )            break;        for(i = 1; i < 500; i++)        {            if(strlen(fib[i]) > strlen(a))            {                flag1 = i;                break;            }            if(strlen(fib[i]) == strlen(a) && strcmp(fib[i], a) == 0)            {                flag1 = i;  flag = 1;                break;            }            if(strlen(fib[i]) == strlen(a) && strcmp(fib[i], a) > 0)            {                flag1 = i;                break;            }        }        for(i = 1; i < 500; i++)        {            if(strlen(fib[i]) > strlen(b))            {                flag2 = i;                break;            }            if(strlen(fib[i]) == strlen(b) && strcmp(fib[i], b) == 0)            {                flag2 = i;                if(flag == 1)                    flag2++;                break;            }            if(strlen(fib[i]) == strlen(b) && strcmp(fib[i], b) > 0)            {                flag2 = i;                break;            }        }        printf("%d\n", flag2 - flag1);    }    return 0;}