How Many Fibs?

How Many Fibs?

点我

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5158 Accepted Submission(s):
2007


[align=left]Problem Description[/align]
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].

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case output on a single line the number
of Fibonacci numbers fi with a <= fi <= b.

[align=left]Sample Input[/align]

10 100
1234567890 9876543210
0 0

[align=left]Sample Output[/align]

5
4

思路很简单，但做得时候思维不严谨小的地方错了好多，一直在debug

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define len 10
int _cmp(char *a,char *b)
{
int i=0,j,count=0;
if(strlen(a)<strlen(b))
return 1;
else if(strlen(a)>strlen(b))
return -1;
else
{
char x[110];
int j=0,len1=strlen(a);
for(i=len1-1;i>=0;i--)    /*倒置，全部都是高位在右，从右往左比较*/
x[j++]=a[i];
x[j]='\0';
i=len1-1;
while(i>=0)
{
if(b[i]>x[i])
return 1;
else if(b[i]==x[i])
i--;
else
{
return -1;
}
}
return 0;
}
}
char f[600][110];
int main()
{
memset(f,'0',sizeof(f));
int c,i,j,count,n;
char a[101],b[101];
f[1][0]='1';
f[2][0]='2';
for(i=3;i<600;i++)
{
c=0;
for(j=0;j<110;j++)
{
f[i][j]=(f[i-2][j]+f[i-1][j]+c-2*'0')%len+'0';
c=(f[i-2][j]+f[i-1][j]+c-2*'0')/len;
}
}
for(i=1;i<600;i++)    /*变成字符串方便实用strlen（）*/
for(j=109;j>=0;j--)
if(f[i][j]!='0')
{
f[i][j+1]='\0';
break;
}
while(cin>>a>>b)
{
if(a[0]=='0'&&b[0]=='0')
break;
count=0;
j=0;
for(i=1;i<=599;i++)
{
if(_cmp(a,f[i])>=0&&_cmp(b,f[i])<=0)
count++;
if(_cmp(b,f[i])>=0)
break;
}
cout<<count<<endl;
}
}