UVa 10183 How Many Fibs? (統計斐波那契數個數&高精度)

10183 - How Many Fibs?

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1124

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 Specification

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<=10100. The numbers a and b are given with no superfluous leading zeros.

Output Specification

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

先打表，再從頭開始枚舉判斷即可。

完整代碼：

/*0.016s*/    #include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  const int maxn = 105;///注意此值的設定要留意中間結果      char numstr[maxn], numstr2[maxn]; ///輸入輸出介面      struct bign  {      int len, s[maxn];          bign()      {          memset(s, 0, sizeof(s));          len = 1;      }          bign(int num)      {          *this = num;      }          bign(const char* num)      {          *this = num;      }          bign operator = (const int num)      {          char s[maxn];          sprintf(s, "%d", num);          *this = s;          return *this;      }          bign operator = (const char* num)      {          len = strlen(num);          for (int i = 0; i < len; i++) s[i] = num[len - i - 1] & 15;          return *this;      }          ///輸出      const char* str() const    {          if (len)          {              for (int i = 0; i < len; i++)                  numstr[i] = '0' + s[len - i - 1];              numstr[len] = '\0';          }          else strcpy(numstr, "0");          return numstr;      }          ///去前置字元為零      void clean()      {          while (len > 1 && !s[len - 1]) len--;      }          ///加      bign operator + (const bign& b) const    {          bign c;          c.len = 0;          for (int i = 0, g = 0; g || i < max(len, b.len); i++)          {              int x = g;              if (i < len) x += s[i];              if (i < b.len) x += b.s[i];              c.s[c.len++] = x % 10;              g = x / 10;          }          return c;      }          ///減      bign operator - (const bign& b) const    {          bign c;          c.len = 0;          for (int i = 0, g = 0; i < len; i++)          {              int x = s[i] - g;              if (i < b.len) x -= b.s[i];              if (x >= 0) g = 0;              else            {                  g = 1;                  x += 10;              }              c.s[c.len++] = x;          }          c.clean();          return c;      }          ///乘      bign operator * (const bign& b) const    {          bign c;          c.len = len + b.len;          for (int i = 0; i < len; i++)              for (int j = 0; j < b.len; j++)                  c.s[i + j] += s[i] * b.s[j];          for (int i = 0; i < c.len - 1; i++)          {              c.s[i + 1] += c.s[i] / 10;              c.s[i] %= 10;          }          c.clean();          return c;      }          ///除      bign operator / (const bign &b) const    {          bign ret, cur = 0;          ret.len = len;          for (int i = len - 1; i >= 0; i--)          {              cur = cur * 10;              cur.s[0] = s[i];              while (cur >= b)              {                  cur -= b;                  ret.s[i]++;              }          }          ret.clean();          return ret;      }          ///模、餘      bign operator % (const bign &b) const    {          bign c = *this / b;          return *this - c * b;      }          bool operator < (const bign& b) const    {          if (len != b.len) return len < b.len;          for (int i = len - 1; i >= 0; i--)              if (s[i] != b.s[i]) return s[i] < b.s[i];          return false;      }          bool operator > (const bign& b) const    {          return b < *this;      }          bool operator <= (const bign& b) const    {          return !(b < *this);      }          bool operator >= (const bign &b) const    {          return !(*this < b);      }          bool operator != (const bign &b) const    {          return b < *this || *this < b;      }          bool operator == (const bign& b) const    {          return !(b < *this) && !(*this < b);      }          bign operator += (const bign &a)      {          *this = *this + a;          return *this;      }          bign operator -= (const bign &a)      {          *this = *this - a;          return *this;      }          bign operator *= (const bign &a)      {          *this = *this * a;          return *this;      }          bign operator /= (const bign &a)      {          *this = *this / a;          return *this;      }          bign operator %= (const bign &a)      {          *this = *this % a;          return *this;      }  };      bign f[500] = {1, 1};      int main()  {      bign a, b;      int i, j;      for (i = 2; i < 500; ++i)          f[i] = f[i - 1] + f[i - 2];      while (scanf("%s%s", numstr, numstr2), a = numstr, b = numstr2, b != 0)      {          i = 1;          while (f[i++] < a)              ;          --i;          j = i;          while (f[j++] <= b)              ;          printf("%d\n", j - i - 1);      }      return 0;  }

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/*0.372s*/    import java.io.*;  import java.util.*;  import java.math.*;      public class Main {      static final int maxn = 500;      static Scanner cin = new Scanner(new BufferedInputStream(System.in));          public static void main(String[] args) {          BigInteger[] f = new BigInteger[maxn];          f[1] = BigInteger.ONE;          f[2] = new BigInteger("2");          for (int i = 3; i < maxn; ++i)              f[i] = f[i - 1].add(f[i - 2]);          while (true) {              BigInteger a = cin.nextBigInteger(), b = cin.nextBigInteger();              if (b.compareTo(BigInteger.ZERO) == 0)                  break;              int i = 1;              while (f[i++].compareTo(a) < 0)                  ;              --i;              int j = i;              while (f[j++].compareTo(b) < 1)                  ;              System.out.println(j - i - 1);          }      }  }