How
Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3804    Accepted Submission(s): 1498 

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

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <vector>

#include <cmath>

#include <iostream>

#include <algorithm>

#include <string>

#include <map>

using namespace std;

const int maxn=500;

const int base=1000000;

const int INF=99999999;

const int MIN=-INF;

char a[maxn][500];

char* add(char*a,char*b)//大数加法

{

    int lena=strlen(a);

    int lenb=strlen(b);

    int i,j,k;

    char c[500];

    if(lena<lenb)

    {

        strcpy(c,a);

        strcpy(a,b);

        strcpy(b,c);

    }

    int A[500]= {0},B[500]= {0};

    for(j=0,i=lena-1; i>=0; i--,j++)

        A[j]=a[i]-‘0‘;

    int lenA=j;

    for(i=lenb-1,j=0; i>=0; i--,j++)

        B[j]=b[i]-‘0‘;

    int lenB=j;

    for(i=0; i<lenB; i++)

        A[i]+=B[i];

    for(i=0; i<500; i++)

    {

        if(A[i]>9)

        {

            A[i+1]+=A[i]/10;

            A[i]%=10;

        }

    }

    for(i=500-1; i>=0; i--)if(A[i]!=0)break;

    for(j=0; i>=0; i--)

        c[j++]=A[i]+‘0‘;

    c[j]=‘\0‘;

    return c;

}

int comper(char*a,char*b)//自定义比较函数 相同是0 大于是1 小于是-1

{

    int lena=strlen(a);

    int lenb=strlen(b);

    if(lena>lenb)

        return 1;

    else if(lena<lenb)

        return -1;

    else if(lena==lenb)

    {

        for(int i=0; i<lena; i++)

            if(a[i]<b[i])

                return -1;

            else if(a[i]>b[i])

                return 1;

    }

    return 0;

}

int main()

{

    int n,m,i,j,k,t;

    a[1][0]=‘1‘;

    a[2][0]=‘2‘;

    for(i=3; i<=maxn; i++)

    {

        strcpy(a[i],add(a[i-1],a[i-2]));

    }

    //for(i=1;i<50;i++)

    //  cout<<a[i]<<endl;
char a1[maxn],b1[maxn];

    while(cin>>a1>>b1)

    {

        int lena1=strlen(a1);

        int lenb1=strlen(b1);

        if(lena1==1&&lenb1==1&&a1[0]==‘0‘&&b1[0]==‘0‘)return 0;

        int cnt=0;

        for(i=1; i<maxn; i++)

        {

            if(comper(a[i],a1)>=0&&comper(a[i],b1)<=0)

                cnt++;

        }

        cout<<cnt<<endl;

    }
return 0;

}

杭电1316