How Many Fibs? 大数加法

发布时间：2020-12-14 03:30:39 所属栏目：大数据来源：网络整理

导读：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

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

求a--b内斐波那契数的个数

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
string str[505],a,b;
string add(string s1,string s2)
{
    string temp;
    int i,j;
    if(s1.length()<s2.length())
    {
        temp=s1;
        s1=s2;
        s2=temp;
    }
    for(int i=s1.length()-1,j=s2.length ()-1;i>=0;i--,j--)
     {
          s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));
          if(s1[i]-'0'>=10)
          {
              s1[i]=char((s1[i]-'0')%10+'0');
              if(i)
                   s1[i-1]++;
              else
                   s1='1'+s1;//一定是这个顺序，1会加到s1前面
          }
     }
    return s1;
}
int main()
{
    int i,j,alen,blen,len,head,end;
    str[1]="1";str[2]="2";
    for(i=3;i<505;i++)
    {
        str[i]=add(str[i-2],str[i-1]);
    }
    while(cin>>a>>b)
    {
        if(a=="0"&&b=="0")
        break;
        alen=a.length();
        blen=b.length();
        for(i=1;i<505;i++)
        {
            len=str[i].length();
            if(len<alen)
            continue;
            if(len==alen&&str[i]>=a)
            {
                head=i;
                break;
            }
            else if(len>alen)
            {
                head=i;
                break;

            }
        }
        for(i=504;i>=1;i--)
        {
            len=str[i].length();
            if(len>blen)
            continue;
            if(len==blen&&str[i]<=b)
            {
                end=i;
                break;
            }
            else if(len<blen)
            {
                end=i;
                break;
            }
        }
        printf("%dn",end-head+1);
    }
    return 0;
}

（编辑：李大同）

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