UVA - 10183 - How Many Fibs? (斐波那契 + 高精度)

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標籤:acm   uva   高精度   數論   fibonacci   


題目傳送:UVA - 10183


思路:高精度就可以了,因為10^100以內的斐波那契數不多,根據公式來看,估計就500多,開個1000的數組足夠啦,實現的話是用的java,注意這裡的斐波那契是從1開始的,我一開始是從0開始的,wa了一下


AC代碼:

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) != -1 && fibo[i].compareTo(b) != 1) {ans ++;}}System.out.println(ans);}}}


















UVA - 10183 - How Many Fibs? (斐波那契 + 高精度)

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