hdu 1316 How Many Fibs?

题目连接

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

How Many Fibs?

Description

Recall the definition of the Fibonacci numbers:
$f_1 := 1$
$f_2 := 2$
$f_n := f_{n-1} + f_{n-2} \ \ (3 \leq n)$

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, $a \leq b \leq 10^{100}$ 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 $f_i$ with $a \leq f_i \leq b. $

SampleInput

10 100

1234567890 9876543210

0 0

SampleOutput

5

4

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cassert>
 #include<cstdio>
 #include<vector>
 #include<string>
 #include<set>
 using std::cin;
 using std::max;
 using std::cout;
 using std::endl;
 using std::string;
 using std::vector;
 using std::istream;
 using std::ostream;
 #define N 510
 #define sz(c) (int)(c).size()
 #define all(c) (c).begin(), (c).end()
 #define iter(c) decltype((c).begin())
 #define cls(arr,val) memset(arr,val,sizeof(arr))
 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
 #define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)
 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
 #define pb(e) push_back(e)
 #define mp(a, b) make_pair(a, b)
 struct BigN {
     typedef unsigned long long ull;
     static const int Max_N = ;
     int len, data[Max_N];
     BigN() { memset(data, , sizeof(data)), len = ; }
     BigN(const int num) {
         memset(data, , sizeof(data));
         *this = num;
     }
     BigN(const char *num) {
         memset(data, , sizeof(data));
         *this = num;
     }
     void clear() { len = , memset(data, , sizeof(data)); }
     BigN& clean(){ while (len >  && !data[len - ]) len--;  return *this; }
     string str() const {
         string res = "";
         for (int i = len - ; ~i; i--) res += (char)(data[i] + '');
         if (res == "") res = "";
         res.reserve();
         return res;
     }
     BigN operator = (const int num) {
         int j = , i = num;
         do data[j++] = i % ; while (i /= );
         len = j;
         return *this;
     }
     BigN operator = (const char *num) {
         len = strlen(num);
         for (int i = ; i < len; i++) data[i] = num[len - i - ] - '';
         return *this;
     }
     BigN operator + (const BigN &x) const {
         BigN res;
         int n = max(len, x.len) + ;
         for (int i = , g = ; i < n; i++) {
             int c = data[i] + x.data[i] + g;
             res.data[res.len++] = c % ;
             g = c / ;
         }
         return res.clean();
     }
     BigN operator * (const BigN &x) const {
         BigN res;
         int n = x.len;
         res.len = n + len;
         for (int i = ; i < len; i++) {
             for (int j = , g = ; j < n; j++) {
                 res.data[i + j] += data[i] * x.data[j];
             }
         }
         for (int i = ; i < res.len - ; i++) {
             res.data[i + ] += res.data[i] / ;
             res.data[i] %= ;
         }
         return res.clean();
     }
     BigN operator * (const int num) const {
         BigN res;
         res.len = len + ;
         for (int i = , g = ; i < len; i++) res.data[i] *= num;
         for (int i = ; i < res.len - ; i++) {
             res.data[i + ] += res.data[i] / ;
             res.data[i] %= ;
         }
         return res.clean();
     }
     BigN operator - (const BigN &x) const {
         assert(x <= *this);
         BigN res;
         for (int i = , g = ; i < len; i++) {
             int c = data[i] - g;
             if (i < x.len) c -= x.data[i];
             if (c >= ) g = ;
             else g = , c += ;
             res.data[res.len++] = c;
         }
         return res.clean();
     }
     BigN operator / (const BigN &x) const {
         BigN res, f = ;
         for (int i = len - ; ~i; i--) {
             f *= ;
             f.data[] = data[i];
             while (f >= x) {
                 f -= x;
                 res.data[i]++;
             }
         }
         res.len = len;
         return res.clean();
     }
     BigN operator % (const BigN &x) {
         BigN res = *this / x;
         res = *this - res * x;
         return res;
     }
     BigN operator += (const BigN &x) { return *this = *this + x; }
     BigN operator *= (const BigN &x) { return *this = *this * x; }
     BigN operator -= (const BigN &x) { return *this = *this - x; }
     BigN operator /= (const BigN &x) { return *this = *this / x; }
     BigN operator %= (const BigN &x) { return *this = *this % x; }
     bool operator <  (const BigN &x) const {
         if (len != x.len) return len < x.len;
         for (int i = len - ; ~i; i--) {
             if (data[i] != x.data[i]) return data[i] < x.data[i];
         }
         return false;
     }
     bool operator >(const BigN &x) const { return x < *this; }
     bool operator<=(const BigN &x) const { return !(x < *this); }
     bool operator>=(const BigN &x) const { return !(*this < x); }
     bool operator!=(const BigN &x) const { return x < *this || *this < x; }
     bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }
     friend istream& operator >> (istream &in, BigN &x) {
         string src;
         in >> src;
         x = src.c_str();
         return in;
     }
     friend ostream& operator << (ostream &out, const BigN &x) {
         out << x.str();
         return out;
     }
 }A[N + ], k1, k2;
 inline void init() {
     A[] = , A[] = ;
     fork(i, , N) A[i] = A[i - ] + A[i - ];
 }
 int main() {
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     init();
     char str1[N], str2[N];
     while (~scanf("%s %s", str1, str2)) {
         if (str1[] == '' && str2[] == '') break;
         int ans = ;
         k1 = str1, k2 = str2;
         fork(i, , N) { if (k1 <= A[i] && A[i] <= k2) ans++; }
         printf("%d\n", ans);
         k1.clear(), k2.clear();
     }
     return ;
 }