hoj1047-How Many Fibs?(高精度fib序列)

给出两个数，求出这两个数之间有多少个fib数, 但是题目给出的数是10^100所以要用大数来处理

用大数相加算法，求出fib[1024]之内的所有数，然后再找出两数之前多少个开始统计

#include <iostream>
#include <string>

using namespace std;

string operator +(const string &a, const string &b)
{
  int i;
  int sum[2001] = {0};
  int len1 = strlen(a.c_str());
  int len2 = strlen(b.c_str());
  for(i = 0; i<len1; i++)
      sum[i] += a.at(len1 - 1 - i) - '0';
  for(i = 0; i<len2; i++)
      sum[i] += b.at(len2 - 1 - i) - '0';
  int max = len1 > len2 ? len1 : len2;
  for(i = 0; i < max; i++)
  {
      if(sum[i] > 9)
      {
          sum[i + 1] += sum[i] / 10;
          sum[i] %= 10;
          if(i == max - 1)
              max++;

      }
  }
  string tmp("");
  for(i = max - 1; i >= 0; i--)
      tmp += (sum[i] + '0');
  return tmp;
}
static string fib[1024];

int main()
{
  string a,b;
  fib[0] = "1";
  fib[1] = "1";
  for(int i = 2; i < 1024; i++)
  {
      fib[i] = fib[i - 1] + fib[i - 2];
  }
  fib[0] = "0";
  int n;
  while(cin >> a >> b &&(a[0]-'0'||b[0]-'0'))
  {
      int i=1,c=0;
      while(1) //找出不小于a的第一个fib数的下标 
      {  
         if(fib[i].length() > a.length()) break;
         else if(fib[i].length() == a.length() && fib[i]>=a) break;
         i++;
      }
      while(1) 
      {
           if(fib[i]==b) {c++;break;}
           if(fib[i].length() > b.length())  break;
           else if(fib[i].length() == b.length() && fib[i]>b) break;
           i++;  
           c++;  
      }
      cout << c << endl;
  }
  return 0;
}