How many Fibs?(高精度斐波那契数) HDU1316

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

思路：好多人用java的BigInteger来做貌似简单，不过思路我觉的太麻烦，斐波那契数玩的不6，就老老实实的高精度比较吧。总的来说就是先把斐波那契数跑出来，自己测试一下会发现600以内足够了，然后每次for循环一遍计数输出就行了。

代码：

#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;

string Add(string a,string b)
{
    string ans;
    int lena = a.length();
    int lenb = b.length();
    if(lena > lenb)
    {
        for(int i=0 ; i<lena-lenb ; i++)
        {
            b = "0" + b;//注意0加在前面 
        }
    }
    else
    {
        for(int i=0 ; i<lenb-lena ; i++)
        {
            a = "0" + a;
        }
    }
    int len = a.length();
    int cf = 0;//进位
    for(int i=len-1 ; i>=0 ; i--)
    {
        int mid = a[i]-'0' + b[i]-'0' + cf;
        cf = 0;//注意进位清零 
        if(mid>9)
        {
            cf = 1;
            mid %= 10;
        }
        ans = char('0'+mid) + ans;
    }
    if(cf)ans = "1" + ans;//注意最后可能有进位 
    return ans;
}

int Compare(string a,string b)
{
    if(a == b)return 0;
    int lena = a.length();
    int lenb = b.length();
    if(lena > lenb)return 1;
    else if(lena < lenb)return -1;
    else if(lena == lenb)
    {
        for(int i=0 ; i<lena ; i++)
        {
            if(a[i] > b[i])return 1;
            else if(a[i] < b[i])return -1;
        }
    }
}

string Fibo[600];

int main()
{
    Fibo[0] = "1";
    Fibo[1] = "2";
    for(int i=2 ; i<600 ; i++)
    {
        Fibo[i] = Add(Fibo[i-1],Fibo[i-2]);
    }
    string a,b;
    while(cin>>a>>b)
    {
        if(a == "0" && b == "0")break;
        if(b == "0")
        {
            printf("0\n");
            continue;
        }
        int sum = 0;
        for(int i=0 ; i<600 ; i++)
        {
            if(Compare(a,Fibo[i]) == 1)continue;
            else if(Compare(b,Fibo[i]) == -1)break;
            sum++;
        }
        printf("%d\n",sum);
    }

    return 0;
}