高精度加二分（二分有坑点）

How many Fibs?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11744		Accepted: 4257

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<=10100. 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
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=550;
const int maxnlen=130;
int f[maxn][maxnlen];
char fi[maxn][maxnlen];
void fibo()
{
f[1][0]=1;
f[2][0]=2;
for(int i=3; i<=500; i++)
for(int j=0; j<=110; j++)
{
f[i][j]+=(f[i-1][j]+f[i-2][j]);
if(f[i][j]>=10)
{
f[i][j+1]=f[i][j]/10;
f[i][j]=f[i][j]%10;
}
}
for(int i=1; i<=500; i++)
{
int j;
for( j=110; j>=0; j--)
if(f[i][j]==0)
continue;
else
break;
int k=0;
for(; j>=0; j--)
fi[i][k++]=f[i][j]+'0';
fi[i][k]='\0';
}
}
int cmp(char *a,char *b)
{
int lena=strlen(a);
int lenb=strlen(b);
if(lena!=lenb)
return lena>lenb?1:-1;
else
return strcmp(a,b);

}
int search(char *a,bool &flag)
{
int low=1;
int high=480;
while(low<=high)
{
int mid=(low+high)/2;
int res=cmp(a,fi[mid]);
if(res==0)
{
flag=1;
return mid;
}
else if(res<0)
high=mid-1;

else
low=mid+1;
}
return high;

}
char a[maxnlen],b[maxnlen];
int main()
{
fibo();
while(~scanf("%s%s",a,b))
{
if(strcmp(a,"0")==0&&strcmp(b,"0")==0)
return 0;
bool flaga =0;
bool flagb=0;
int left=search(a,flaga);
int right=search(b,flagb);
if(flaga)
{
printf("%d\n",right-left+1);

}

else
printf("%d\n",right-left);

}
return 0;
}