n-n rmi mina -- leading else ring tput length

題目鏈接：

http://acm.hdu.edu.cn/showproblem.php?pid=1316

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 代碼如下：

#include<bits/stdc++.h>
using namespace std;
string str[500];
string add(string str1,string str2)//大數相加
{
    
 
int l1=str1.length();
    int l2=str2.length();
    if(l1>l2)
    {
        for(int i=0;i<l1-l2;i++)
        {
            str2="0"+str2;
        }
    }else if(l1<l2)
    {
        for(int i=0;i<l2-l1;i++)
        {
            str1="0"+str1;
        }
    }
    l1=str1.length();
    string
 
 str3;
    int c=0;
    for(int i=l1-1;i>=0;i--)
    {
        int temp=str1[i]-‘0‘+str2[i]-‘0‘+c;
        c=temp/10;
        temp=temp%10;
        str3=char(temp+‘0‘)+str3;
    }
    if(c!=0)
    {
        str3=char(c+‘0‘)+str3;
    }
    return str3;
}
int compare(string str1,string str2)//str1大於等於str2，返回1
{
    int l1=str1.length();
    int l2=str2.length();
    if(l1>l2)
    {
        return 1;
    }else if(l1<l2)
    {
        return 0;
    }else
    {
        for(int i=0;i<l1;i++)
        {
            if(str1[i]>str2[i])
            {
                return 1;
            }else if(str1[i]==str2[i])
            {
                continue;
            }else
            {
                return 0;
            }
        }
    }
    return 1;
}
int f(string str1,string str2)//找出兩個大數中間的斐波那契數的個數，包括邊界
{
    int l1=str1.length(),l2=str2.length(),index1,index2;
    for(int i=0;i<=500;i++)
    {
        int k=str[i].length();
        if(k<l1)
            continue;
        else
        {
            index1=i;
            break;
        }
    }
    for(int i=499;i>=0;i--)
    {
        int k=str[i].length();
        if(k>l2)
            continue;
        else
        {
            index2=i;
            break;
        }
    }
    int r=0;
    for(int i=index1;i<=index2;i++)
    {
        if(compare(str[i],str1)==1&&compare(str2,str[i])==1)
        {
            r++;
        }
    }
    return r;
}
int main()
{
    string str1,str2;
    str[0]="1";
    str[1]="2";
    for(int i=2;i<500;i++)//先求出需要的斐波那契數列，第500個斐波那契數大於10的1000次方
    {
        str[i]=add(str[i-1],str[i-2]);
    }
    while(cin>>str1>>str2)
    {
        if(str1=="0"&&str2=="0")
            break;
        int r=f(str1,str2);
        printf("%d\n",r);
    }
    return 0;
}

HDU 1316 （斐波那契數列，大數相加，大數比較大小）