hdu 1316

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2282    Accepted Submission(s): 926

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

 

解题思路：大整数相加，存放到2维数组里，然后二分查找就可以了。本题对大整数是一个很好的练习。

 

#include <iostream>
#include <cstdlib>
#include <cstring>
#define M 105
using namespace std;
char data[1000][M+2];
char a[M+2],b[M+2];
int cmp_ab(char *s1,char *s2)
{
    for(int k=0;k<=105;k++)
      {
         if (k==105)
             return s1[k]-s2[k];
          if (s1[k]!=s2[k])  return s1[k]-s2[k];

      }

}


int find1(int i,char *x)
{
    int low=0,high=i,mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        int value=cmp_ab(x,data[mid]);
        if (value>0)  low=mid+1;
        if (value==0) return mid-1;
        if (value<0)  high=mid-1;
    }
     return high; //这时候high小于low
}

int find2(int i,char *x)
{
    int low=0,high=i,mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        int value=cmp_ab(x,data[mid]);
        if (value>0)  low=mid+1;
        if (value==0) return mid+1;
        if (value<0)  high=mid-1;
    }
     return low; //这时候 low大于high
}



int main()
{
  int i,j,p;
  //memset(data,0,sizeof(data));
  data[0][105]=1;
  data[1][105]=2;
  i=2;p=105;
  while(data[i-1][5]<=1)
  {
    for(j=105;j>=p;j--)
    data[i][j]=data[i-1][j]+data[i-2][j];

    for(j=105;j>=p;j--)
    {
         int c=data[i][j]/10;
         if (c>=1)
         {
             data[i][j]=data[i][j]%10;
             data[i][j-1]=data[i][j-1]+c;
         }

    }
     if (data[i][p-1]>0) p--;
    i++;
  }

    while(cin>>a>>b)
{
    if (a[0]=='0'&&b[0]=='0') break;
     int len1=strlen(a)-1;
     int len2=strlen(b)-1;
     int k;
     for(int d=len1,k=105;d>=0;d--,k--)
        {
             a[k]=a[d]-'0';
             a[d]=0;
        }
      for(int d=len2,k=105;d>=0;d--,k--)
        {
             b[k]=b[d]-'0';
             b[d]=0;
        }
    /*for(int i=0;i<=105;i++)
    cout<<(int)b[i];
    cout<<endl;
    */
    int lt=find1(i-1,a);
    int rt=find2(i-1,b);
    //cout<<"lt="<<lt<<endl;
    //cout<<"rt="<<rt<<endl;
   //cout<<lt<<endl;

    cout<<rt-lt-1<<endl;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));

}

    return 0;
}