How Many Fibs? hdu1316

 

//大整数的加法，花了我两个小时，一定要小心仔细才行呀;
//笨（本）人代码如下
#include<stdio.h>
#include<string.h>
int fib[510][120];//the first element is the array length
int a[120],b[120];

void change(int *x)//right
{
    int i,temp,len;
    len=x[0];
    for(i=1;i<=len/2;i++)
    {
        temp=x[i];x[i]=x[len+1-i];x[len+1-i]=temp;
    }
}

void countfib()// right
{
    int k,i,len1,len2,len,carry;
    memset(fib,0,sizeof(fib));
    fib[1][0]=fib[2][0]=1;
    fib[1][1]=1;fib[2][1]=2;
    for(k=3;k<510;k++)//compute the fib ,is the most important and most difficult of there 
    {//打表算出在其范围内所有的fibs数
        len1=fib[k-2][0];len2=fib[k-1][0];
        len=len1>len2?len1:len2;
        carry=0;
        change(fib[k-2]);//算前将其反置，便于计算，第0个数保存其长度。
        change(fib[k-1]);
        for(i=1;i<=len+1;i++)
        {
            fib[k][i]=fib[k-2][i]+fib[k-1][i]+carry;
            if(fib[k][i]>=10)
            {
                fib[k][i]-=10;
                carry=1;
            }
            else
                carry=0;
        }
        if(fib[k][len+1]==0)
            fib[k][0]=len;
        else
            fib[k][0]=len+1;
        change(fib[k]);
        change(fib[k-1]);//一定要记得算完后所有的数都要转换回来。
        change(fib[k-2]);
    }

}

int compare(int *x,int *y)//x>y 0, x==y 2, x<y 1 right
{
    int len1,len2,i;
    len1=x[0];len2=y[0];
    if(len1>len2)
        return 0;
    else if(len1<len2)
        return 1;
    else
    {
        for(i=1;i<=len1;i++)
        {
            if(x[i]>y[i])
                return 0;
            else if(x[i]<y[i])
                return 1;
        }
        return 2;
    }
}

int main()
{
    int i,j,flag,count;
    int start,end;
    char s1[120],s2[120];
    countfib();
    while(1)
    {
        scanf("%s",s1);
        scanf("%s",s2);
        if(s1[0]=='0' && s2[0]=='0')
            break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0;s1[i];i++)
            a[i+1]=s1[i]-'0';
        a[0]=i;
        for(i=0;s2[i];i++)
            b[i+1]=s2[i]-'0';
        b[0]=i;
        count=0;

        for(i=1;;i++)
        {
            flag=compare(a,fib[i]);
            if(flag>0)
            { start=i;break;}
        }
        for(i=1;;i++)
        {
            flag=compare(b,fib[i]);
            if(flag==1)
            {     end=i-1;break;}
            else if(flag==2)
            {   end=i;break;}
        }
        count=end-start+1;
        printf("%d\n",count);
    /*    printf("start=%d end=%d count=%d\n",start,end,count);//有时打印中间值查看错误是一种很好的习惯。
        for(i=start;i<=end;i++)
        {
            for(j=1;j<=fib[i][0];j++)
                printf("%d",fib[i][j]);
            printf("\n");
        }*/
    }
    return 0;
}

How Many Fibs?

Problem Description hdu1316 

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