题目大意:给定一个范围a,b为再次范围内有多少的斐波那契额数.
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思路:直接上java搞,注意在java中的a==0&&b==0程序停止的写法.
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import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static BigInteger[] ans;
public static void main(String[] args)
{
ans=new BigInteger[1010];
Scanner cin=new Scanner(System.in);
int n,i;
ans[1]=BigInteger.valueOf(1);
ans[2]=BigInteger.valueOf(2);
for(i=3;i<=1000;i++)
{
ans[i]=ans[i-1].add(ans[i-2]);
}
BigInteger a,b;
while(true)
{
a=cin.nextBigInteger();
b=cin.nextBigInteger();
if(a.compareTo(b)==0&&a.compareTo(BigInteger.valueOf(0))==0)
break;
int tmp;
i=1;tmp=0;
while(ans[i].compareTo(a)<0)
i++;
while(ans[i].compareTo(b)<=0&&ans[i].compareTo(a)>=0)
{
tmp++;
i++;
}
System.out.println(tmp);
}
}
}
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