How Many Fibs?

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1075    Accepted Submission(s): 454

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

 

 

 

大整数的加法，花了我两个小时，一定要小心仔细才行呀;

笨（本）人代码如下

#include<stdio.h>
#include<string.h>

int fib[510][120];//the first element is the array length

int a[120],b[120];

void change(int *x)//right
{
 int i,temp,len;
 len=x[0];
 for(i=1;i<=len/2;i++)
 {
  temp=x[i];x[i]=x[len+1-i];x[len+1-i]=temp;
 }
}

void countfib()// right
{
 int k,i,len1,len2,len,carry;
 memset(fib,0,sizeof(fib));
 fib[1][0]=fib[2][0]=1;
 fib[1][1]=1;fib[2][1]=2;
 for(k=3;k<510;k++)//compute the fib ,is the most important and most difficult of there
 {
  len1=fib[k-2][0];len2=fib[k-1][0];
  len=len1>len2?len1:len2;
  carry=0;
  change(fib[k-2]);
  change(fib[k-1]);
  for(i=1;i<=len+1;i++)
  {
   fib[k][i]=fib[k-2][i]+fib[k-1][i]+carry;
   if(fib[k][i]>=10)
   {
    fib[k][i]-=10;
    carry=1;
   }
   else
    carry=0;
  }
  if(fib[k][len+1]==0)
   fib[k][0]=len;
  else
   fib[k][0]=len+1;
  change(fib[k]);
  change(fib[k-1]);
  change(fib[k-2]);
 }

}

int compare(int *x,int *y)//x>y 0, x==y 2, x<y 1 right
{
 int len1,len2,i;
 len1=x[0];len2=y[0];
 if(len1>len2)
  return 0;
 else if(len1<len2)
  return 1;
 else
 {
  for(i=1;i<=len1;i++)
  {
   if(x[i]>y[i])
    return 0;
   else if(x[i]<y[i])
    return 1;
  }
  return 2;
 }
}

int main()
{
 int i,j,flag,count;
 int start,end;
 char s1[120],s2[120];
 countfib();
 while(1)
 {
  scanf("%s",s1);
  scanf("%s",s2);
  if(s1[0]=='0' && s2[0]=='0')
   break;
  memset(a,0,sizeof(a));
  memset(b,0,sizeof(b));
  for(i=0;s1[i];i++)
   a[i+1]=s1[i]-'0';
  a[0]=i;
  for(i=0;s2[i];i++)
   b[i+1]=s2[i]-'0';
  b[0]=i;
  count=0;

  for(i=1;;i++)
  {
   flag=compare(a,fib[i]);
   if(flag>0)
   { start=i;break;}
  }
  for(i=1;;i++)
  {
   flag=compare(b,fib[i]);
   if(flag==1)
   {  end=i-1;break;}
   else if(flag==2)
   {   end=i;break;}
  }
  count=end-start+1;
  printf("%d/n",count);
 /* printf("start=%d end=%d count=%d/n",start,end,count);
  for(i=start;i<=end;i++)
  {
   for(j=1;j<=fib[i][0];j++)
    printf("%d",fib[i][j]);
   printf("/n");
  }*/
 }
 return 0;
}