How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4235    Accepted Submission(s): 1669

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

题意：

给定范围[a,b]（a<=b<=10^100），求在这范围内的斐波那契数有多少个。

思路：

在我印象中，1000项的斐波那契数好像就有200多位了，枚举1---1000项的斐波那契数放到数组里，然后strcmp求在[a,b]之间的斐波那契的个数。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <iostream>
 6 using namespace std;
 7
 8 char a[110], b[110];
 9 char f[1005][500];
10 int f1[500];
11 int f2[500];
12
13 void init_f(){                 //高精度加法初始化斐波那契数f[]
14     strcpy(f[0],"1");
15     strcpy(f[1],"1");
16     int i, j, k, l, r;
17     for(i=2;i<=1000;i++){
18         int n1=strlen(f[i-2]);
19         int n2=strlen(f[i-1]);
20         memset(f1,0,sizeof(f1));
21         memset(f2,0,sizeof(f2));
22         for(j=n1-1;j>=0;j--) f1[j]=f[i-2][n1-j-1]-‘0‘;
23         for(j=n2-1;j>=0;j--) f2[j]=f[i-1][n2-j-1]-‘0‘;
24         int c=0;
25         while(j<=max(n1,n2)){
26             f1[j]=f1[j]+f2[j]+c;
27             if(f1[j]>=10) {
28                 f1[j]-=10;c=1;
29             }
30             else c=0;
31             j++;
32         }
33         k=max(n1,n2);
34         while(!f1[k]) k--;          //去掉后面多余的0
35         for(j=k;j>=0;j--) f[i][j]=f1[k-j]+‘0‘;
36
37     }
38 }
39
40
41 main()
42 {
43     init_f();
44     int ans;
45     while(scanf("%s%s",a,b)!=EOF&&(strcmp(a,"0")||strcmp(b,"0"))){
46         int n1=strlen(a), n2=strlen(b);
47         int flag;
48         if(n1==n2) flag=1;
49         else flag=0;
50         ans=0;
51         for(int i=1;i<=1000;i++){
52             int n=strlen(f[i]);
53             if(flag){                                 //当a和b的长度一样时
54                 if(n==n1&&strcmp(f[i],a)>=0&&strcmp(f[i],b)<=0) ans++;
55                 if(n>n1||n==n1&&strcmp(f[i],b)>0) break;
56                 continue;
57             }
58             if(n>n2||n==n2&&strcmp(f[i],b)>0) break;      //当a和b的长度不一样时
59             if(n>n1&&n<n2) {
60                 ans++;continue;
61             }
62             if(n==n1&&strcmp(f[i],a)>=0) ans++;
63             if(n==n2&&strcmp(f[i],b)<=0) ans++;
64         }
65         printf("%d\n",ans);
66     }
67 }

HDU 1316 斐波那契数列+高精度,布布扣,bubuko.com