How Many Fibs?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5638 Accepted Submission(s):
2183
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].
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
Source
Recommend
大数问题，不是很难，不过处理起来有点繁琐，需要仔细一点。
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。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 char ch[1005][500]; 6 int f1[500]; 7 int f2[500]; 8 void init() //高精度加法初始化斐波那契数ch[] 9 { 10 int i,j; 11 strcpy(ch[0],"1"); 12 strcpy(ch[1],"1"); 13 for(i=2; i<=1000; i++) 14 { 15 int n1=strlen(ch[i-2]); 16 int n2=strlen(ch[i-1]); 17 memset(f1,0,sizeof(f1)); 18 memset(f2,0,sizeof(f2)); 19 for(j=n1-1; j>=0; j--) f1[j]=ch[i-2][n1-j-1]-'0'; 20 for(j=n2-1; j>=0; j--) f2[j]=ch[i-1][n2-j-1]-'0'; 21 int c=0; 22 j=0; 23 while(j<=max(n1,n2)) 24 { 25 f1[j]=f1[j]+f2[j]+c; 26 if(f1[j]>=10) 27 { 28 f1[j]-=10; 29 c=1; 30 } 31 else c=0; 32 j++; 33 } 34 int w=max(n1,n2); 35 while(!f1[w]) w--; //去掉后面多余的0 36 for(j=w; j>=0; j--) 37 ch[i][j]=f1[w-j]+'0'; 38 } 39 } 40 int main() 41 { 42 init(); 43 int k,flag,i,j; 44 char str1[105],str2[105]; 45 while(~scanf("%s%s",str1,str2)&&(strcmp(str1,"0")||strcmp(str2,"0"))) 46 { 47 int n1=strlen(str1),n2=strlen(str2); 48 if(n1==n2) flag=1; 49 else flag=0; 50 k=0; 51 for(i=1; i<=1000; i++) 52 { 53 int n=strlen(ch[i]); 54 if(flag) //当str1和str2的长度一样时 55 { 56 if(n==n1&&strcmp(ch[i],str1)>=0&&strcmp(ch[i],str2)<=0) k++; 57 if(n>n1||n==n1&&strcmp(ch[i],str2)>0) break; 58 continue; 59 } 60 if(n>n2||n==n2&&strcmp(ch[i],str2)>0) break; //当str1和str2的长度不一样时 61 if(n>n1&&n<n2) 62 { 63 k++; 64 continue; 65 } 66 if(n==n1&&strcmp(ch[i],str1)>=0) k++; 67 if(n==n2&&strcmp(ch[i],str2)<=0) k++; 68 } 69 printf("%d ",k); 70 } 71 return 0; 72 }