YT15-HDU-How many fibs(大数相加法)

发布时间：2020-12-14 02:45:21 所属栏目：大数据来源：网络整理

导读：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 s

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

10 100
1234567890 9876543210
0 0

Sample Output

5
4

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：

（编辑：李大同）

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