HDU1316:How Many Fibs?
Problem DescriptionRecall 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].

InputThe 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.

OutputFor each test case output on a single line the number of Fibonacci numbers fi with a <= fi <= b.

Sample Input

10 1001234567890 98765432100 0

54

//题意，在给定的范围中有几个斐波数，注意，提交的语言选G++，选C++会提示CE

#include<iostream>#include<cstring>#include<cstdio>using namespace std;string add(string s1,string s2){     if(s1.length()<s2.length())     {          string temp=s1;          s1=s2;          s2=temp;     }     for(int i=s1.length()-1,j=s2.length ()-1;i>=0;i--,j--)     {          s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));          if(s1[i]-'0'>=10)          {              s1[i]=char((s1[i]-'0')%10+'0');              if(i)                   s1[i-1]++;              else                   s1='1'+s1;          }     }     return s1;}string str[1005],a,b;int main(){    int i;    str[1] = "1";    str[2] = "2";    for(i = 3;i<1005;i++)    {        str[i] = add(str[i-2],str[i-1]);        //printf("%s\n",str[i]);    }    while(cin >> a >> b)    {        if(a=="0" && b == "0")        break;        int head,end,cnt = 0,len_a,len_b,len;        len_a = a.length();        len_b = b.length();        for(i = 1;i<1005;i++)        {            len = str[i].length();            if(len<len_a)            continue;            else if(len == len_a && str[i]>=a)            {                head = i;                break;            }            else if(len > len_a)            {                head = i;                break;            }        }        for(i = 1004;i>=1;i--)        {            len = str[i].length();            if(len>len_b)            continue;            else if(len == len_b && str[i]<=b)            {                end = i;                break;            }            else if(len < len_b)            {                end = i;                break;            }        }        printf("%d\n",end-head+1);    }    return 0;}