YT15-HDU-How many fibs(大数相加法)

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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 1001234567890 98765432100 0

Sample Output

代码如下：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;string add(string x,string y)                            //大数相加法{    int i,temp,m=0;    string Fib;    int lenx=x.length();    int leny=y.length();                                 //返回字符串x，y的长度    if (lenx<leny)                                       //如果x比y短    {        for (i=1; i<=leny-lenx; i++)                     //在x前面补0至与y相同的长度            x="0"+x;    }    else                                                 //反之    {        for (i=1; i<=lenx-leny; i++)                     //在y前面补0至与x相同的长度            y="0"+y;    }    lenx = x.length();                                   //x长度发生变化，需要重新返回长度    for (i=lenx-1; i>=0; i--)    {        temp=x[i]-'0'+y[i]-'0'+m;                        //相同位置的数字相加并加上后一位进的数        m=temp/10;                                       //前一位将要加上m        temp%=10;                                        //取余        Fib=char (temp+'0')+Fib;                         //将这一位的数放进新的字符串    }    if (m!=0)        Fib=char (m+'0')+Fib;                            //如果最高位的数大于10继续往前进    return Fib;                                          //返回字符串}int main(){    string fibs[1005],a,b;    int i;    fibs[1]="1";    fibs[2]="2";    for (i=3; i<1005; i++)    {        fibs[i]=add(fibs[i-2],fibs[i-1]);                //将前1005个斐波那契数用数组统计。    }    while(cin >> a >> b)    {        if(a=="0" && b == "0")            break;        int head,end,len_a,len_b,len;        len_a = a.length();                             //返回区间最小值的长度        len_b = b.length();                             //返回区间最大值的长度        //找到区间内最小的斐波那契数的位置        for(i = 1; i<1005; i++)        {            len = fibs[i].length();                     //返回某一斐波那契数的长度            if(len<len_a)                continue;            else if(len == len_a && fibs[i]>=a)         //长度相等时比较字符串的大小            {                head = i;                break;            }            else if(len > len_a)                                   {                head = i;                break;            }        }                    //找到区间内最大的斐波那契数的位置                                         for(i = 1004; i>=1; i--)        {            len = fibs[i].length();            if(len>len_b)                continue;            else if(len == len_b && fibs[i]<=b)            {                end = i;                break;            }            else if(len < len_b)            {                end = i;                break;            }        }        cout<<end-head+1<<endl;                   //首尾相减+1    }    return 0;}

运行结果：

1 0