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

  1 #include<algorithm>

  2 #include<iostream>

  3 #include<cstdlib>

  4 #include<cstring>

  5 #include<cassert>

  6 #include<cstdio>

  7 #include<vector>

  8 #include<string>

  9 #include<set>

 10 using std::cin;

 11 using std::max;

 12 using std::cout;

 13 using std::endl;

 14 using std::string;

 15 using std::vector;

 16 using std::istream;

 17 using std::ostream;

 18 #define N 510

 19 #define sz(c) (int)(c).size()

 20 #define all(c) (c).begin(), (c).end()

 21 #define iter(c) decltype((c).begin())

 22 #define cls(arr,val) memset(arr,val,sizeof(arr))

 23 #define cpresent(c, e) (find(all(c), (e)) != (c).end())

 24 #define rep(i, n) for (int i = 0; i < (int)(n); i++)

 25 #define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)

 26 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)

 27 #define pb(e) push_back(e)

 28 #define mp(a, b) make_pair(a, b)

 29 struct BigN {

 30     typedef unsigned long long ull;

 31     static const int Max_N = 2010;

 32     int len, data[Max_N];

 33     BigN() { memset(data, 0, sizeof(data)), len = 0; }

 34     BigN(const int num) {

 35         memset(data, 0, sizeof(data));

 36         *this = num;

 37     }

 38     BigN(const char *num) {

 39         memset(data, 0, sizeof(data));

 40         *this = num;

 41     }

 42     void clear() { len = 0, memset(data, 0, sizeof(data)); }

 43     BigN& clean(){ while (len > 1 && !data[len - 1]) len--;  return *this; }

 44     string str() const {

 45         string res = "";

 46         for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0');

 47         if (res == "") res = "0";

 48         res.reserve();

 49         return res;

 50     }

 51     BigN operator = (const int num) {

 52         int j = 0, i = num;

 53         do data[j++] = i % 10; while (i /= 10);

 54         len = j;

 55         return *this;

 56     }

 57     BigN operator = (const char *num) {

 58         len = strlen(num);

 59         for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0';

 60         return *this;

 61     }

 62     BigN operator + (const BigN &x) const {

 63         BigN res;

 64         int n = max(len, x.len) + 1;

 65         for (int i = 0, g = 0; i < n; i++) {

 66             int c = data[i] + x.data[i] + g;

 67             res.data[res.len++] = c % 10;

 68             g = c / 10;

 69         }

 70         return res.clean();

 71     }

 72     BigN operator * (const BigN &x) const {

 73         BigN res;

 74         int n = x.len;

 75         res.len = n + len;

 76         for (int i = 0; i < len; i++) {

 77             for (int j = 0, g = 0; j < n; j++) {

 78                 res.data[i + j] += data[i] * x.data[j];

 79             }

 80         }

 81         for (int i = 0; i < res.len - 1; i++) {

 82             res.data[i + 1] += res.data[i] / 10;

 83             res.data[i] %= 10;

 84         }

 85         return res.clean();

 86     }

 87     BigN operator * (const int num) const {

 88         BigN res;

 89         res.len = len + 1;

 90         for (int i = 0, g = 0; i < len; i++) res.data[i] *= num;

 91         for (int i = 0; i < res.len - 1; i++) {

 92             res.data[i + 1] += res.data[i] / 10;

 93             res.data[i] %= 10;

 94         }

 95         return res.clean();

 96     }

 97     BigN operator - (const BigN &x) const {

 98         assert(x <= *this);

 99         BigN res;

100         for (int i = 0, g = 0; i < len; i++) {

101             int c = data[i] - g;

102             if (i < x.len) c -= x.data[i];

103             if (c >= 0) g = 0;

104             else g = 1, c += 10;

105             res.data[res.len++] = c;

106         }

107         return res.clean();

108     }

109     BigN operator / (const BigN &x) const {

110         BigN res, f = 0;

111         for (int i = len - 1; ~i; i--) {

112             f *= 10;

113             f.data[0] = data[i];

114             while (f >= x) {

115                 f -= x;

116                 res.data[i]++;

117             }

118         }

119         res.len = len;

120         return res.clean();

121     }

122     BigN operator % (const BigN &x) {

123         BigN res = *this / x;

124         res = *this - res * x;

125         return res;

126     }

127     BigN operator += (const BigN &x) { return *this = *this + x; }

128     BigN operator *= (const BigN &x) { return *this = *this * x; }

129     BigN operator -= (const BigN &x) { return *this = *this - x; }

130     BigN operator /= (const BigN &x) { return *this = *this / x; }

131     BigN operator %= (const BigN &x) { return *this = *this % x; }

132     bool operator <  (const BigN &x) const {

133         if (len != x.len) return len < x.len;

134         for (int i = len - 1; ~i; i--) {

135             if (data[i] != x.data[i]) return data[i] < x.data[i];

136         }

137         return false;

138     }

139     bool operator >(const BigN &x) const { return x < *this; }

140     bool operator<=(const BigN &x) const { return !(x < *this); }

141     bool operator>=(const BigN &x) const { return !(*this < x); }

142     bool operator!=(const BigN &x) const { return x < *this || *this < x; }

143     bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }

144     friend istream& operator >> (istream &in, BigN &x) {

145         string src;

146         in >> src;

147         x = src.c_str();

148         return in;

149     }

150     friend ostream& operator << (ostream &out, const BigN &x) {

151         out << x.str();

152         return out;

153     }

154 }A[N + 1], k1, k2;

155 inline void init() {

156     A[1] = 1, A[2] = 2;

157     fork(i, 3, N) A[i] = A[i - 1] + A[i - 2];

158 }

159 int main() {

160 #ifdef LOCAL

161     freopen("in.txt", "r", stdin);

162     freopen("out.txt", "w+", stdout);

163 #endif

164     init();

165     char str1[N], str2[N];

166     while (~scanf("%s %s", str1, str2)) {

167         if (str1[0] == '0' && str2[0] == '0') break;

168         int ans = 0;

169         k1 = str1, k2 = str2;

170         fork(i, 1, N) { if (k1 <= A[i] && A[i] <= k2) ans++; }

171         printf("%d\n", ans);

172         k1.clear(), k2.clear();

173     }

174     return 0;

175 }

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