题目传送：UVA - 10183
思路：高精度就能够了，由于10^100之内的斐波那契数不多，根据公式来看，估计就500多，开个1000的数组足够啦，实现的话是用的java，注意这里的斐波那契是从1开始的，我1开始是从0开始的，wa了1下
AC代码：
import java.util.Scanner;
import java.math.BigInteger;
public class Main {
public static void main(String args[]) {
Scanner cin = new Scanner(System.in);
BigInteger a, b;
BigInteger[] fibo = new BigInteger[1005];
fibo[0] = new BigInteger("1");
fibo[1] = new BigInteger("2");
for(int i = 2; i < 1005; i ++) {
fibo[i] = fibo[i - 2].add(fibo[i - 1]);
}
while(true) {
a = cin.nextBigInteger();
b = cin.nextBigInteger();
if(a.compareTo(BigInteger.ZERO) == 0 && b.compareTo(BigInteger.ZERO) == 0) {
break;
}
int ans = 0;
for(int i = 0; i < 1005; i ++) {
if(fibo[i].compareTo(a) != ⑴ && fibo[i].compareTo(b) != 1) {
ans ++;
}
}
System.out.println(ans);
}
}
}