// [4/14/2014 Sjm]
/*
考虑好边界处理，即可 AC 。
(
虽然题目所给数据值很大，但无需优化，亦可水过。。。
下面附 Java 的 AC 代码，第一次用 Java 交代码，
很久不用 Java,照着文档，调试很长时间。。。。。
不得不感叹 Java 对于大数处理的便捷与高效，自己写的C++代码效率太低了。
）
*/
| Accepted |
1316 |
890MS |
276K |
1742 B |
C++ |
1 //(1) C++ 代码
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <cstring>
6 #include <string>
7 #include <cstring>
8 #include <algorithm>
9 using namespace std;
10 string str_1, str_2;
11
12 bool Cmp(string str1, string str2)
13 {
14 int len1 = str1.size(), len2 = str2.size();
15 if (len1 > len2) return true;
16 if (len1 < len2) return false;
17 for (int i = 0; i < len1; i++) {
18 if (str1[i] == str2[i]) continue;
19 if ((str1[i] - '0') >(str2[i] - '0')) return true;
20 else return false;
21 }
22 return true;
23 }
24
25 bool myJudge(string str)
26 {
27 if (Cmp(str, str_1) && Cmp(str_2, str)) return true;
28 else return false;
29 }
30
31 string myAdd(string str1, string str2)
32 {
33 string str = "";
34 int temp = 0, len1 = str1.size() - 1, len2 = str2.size() - 1;
35 while (len1 != -1 && len2 != -1) {
36 temp = (str1[len1--] - '0') + (str2[len2--] - '0') + temp;
37 str = char('0' + temp % 10) + str;
38 temp /= 10;
39 }
40 while (len1 != -1) {
41 temp = (str1[len1--] - '0') + temp;
42 str = char('0' + temp % 10) + str;
43 temp /= 10;
44 }
45 while (len2 != -1) {
46 temp = (str2[len2--] - '0') + temp;
47 str = char('0' + temp % 10) + str;
48 temp /= 10;
49 }
50 if (temp) str = char(temp + '0') + str;
51 return str;
52 }
53
54 int main()
55 {
56 //freopen("input.txt", "r", stdin);
57 //freopen("output.txt", "w", stdout);
58 while (cin >> str_1 >> str_2 && (str_1 != "0" || str_2 != "0"))
59 {
60 int ans = 0;
61 string str, str1 = "1", str2 = "2";
62 if (!Cmp(str_2, str1)) {
63 printf("%d
", ans);
64 continue;
65 }
66 if (myJudge(str1)) ans++;
67 if (!Cmp(str_2, str2)) {
68 printf("%d
", ans);
69 continue;
70 }
71 if (myJudge(str2)) ans++;
72 str = myAdd(str1, str2);
73 while (Cmp(str_2, str)){
74 if (myJudge(str))
75 ans++;
76 str1 = str2;
77 str2 = str;
78 str = myAdd(str1, str2);
79 }
80 printf("%d
", ans);
81 }
82 return 0;
83 }
| Accepted |
1316 |
203MS |
5680K |
884 B |
Java |
1 // Java 代码
2 import java.util.*;
3 import java.math.*;
4 import java.io.*;
5
6 public class Main {
7 public static void main(String[] agrs) {
8 BigInteger[]fib = new BigInteger[500];
9 fib[1] = new BigInteger("1");
10 fib[2] = new BigInteger("2");
11 for (int i = 3; i<500; i++) {
12 fib[i] = fib[i - 2].add(fib[i - 1]);
13
14 // 用于判断 10^100 边界值
15 //String str = fib[i].toString();
16 //if (100 < str.length()) {
17 //System.out.println(i);
18 //}
19 }
20 Scanner input = new Scanner(new BufferedInputStream(System.in));
21 BigInteger mydata1, mydata2;
22 while (input.hasNextBigInteger()){
23 mydata1 = input.nextBigInteger();
24 mydata2 = input.nextBigInteger();
25 if (mydata1.compareTo(BigInteger.valueOf(0)) == 0 && mydata2.compareTo(BigInteger.valueOf(0)) == 0) {
26 break;
27 }
28 int pos = 1, ans = 0;
29 while (fib[pos].compareTo(mydata2) != 1) {
30 if ((fib[pos].compareTo(mydata1) != -1) && (mydata2.compareTo(fib[pos]) != -1)) {
31 ans++;
32 }
33 pos++;
34 }
35 System.out.println(ans);
36 }
37 }
38 }