UVa 10198 Counting：組合數學&高精度

10198 - Counting

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1139

The Problem

Gustavo knows how to count, but he is now learning how write numbers. As he is a very good student, he already learned 1, 2, 3 and 4. But he didn't realize yet that 4 is different than 1, so he thinks that 4 is another way to write 1. Besides that, he is having fun with a little game he created himself: he make numbers (with those four digits) and sum their values. For instance:

132 = 1 + 3 + 2 = 6112314 = 1 + 1 + 2 + 3 + 1 + 1 = 9 (remember that Gustavo thinks that 4 = 1)

After making a lot of numbers in this way, Gustavo now wants to know how much numbers he can create such that their sum is a number n. For instance, for n = 2 he noticed that he can make 5 numbers: 11, 14, 41, 44 and 2 (he knows how to count them up, but he doesn't know how to write five). However, he can't figure it out for n greater than 2. So, he asked you to help him.

The Input

Input will consist on an arbitrary number of sets. Each set will consist on an integer n such that 1 <= n <= 1000. You must read until you reach the end of file.

The Output

For each number read, you must output another number (on a line alone) stating how much numbers Gustavo can make such that the sum of their digits is equal to the given number.

Sample Input

23

Sample Output

513

思路：用f[i]表示n=i時的答案，則考慮末位元字，如果選擇1的話，那麼一共有f[i-1]個數，如果是2的話，一共有f[i-2]個數，3有f[i-3]個數，4有f[i-1]個數。

所以：f[i]=2*f[i-1]+f[i-2]+f[i-3]   (f[1]=2,f[2]=5,f[3]=13)

完整代碼：

/*0.032s*/    #include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  const int maxn = 410;      char numstr[maxn];      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  ///本文URL地址：http://www.bianceng.cn/Programming/sjjg/201410/45360.htm    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 &a) const    {          return *this > a || *this < a;      }          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;      }  } f[1010];      int main(void)  {      f[1] = 2, f[2] = 5, f[3] = 13;      for (int i = 4; i <= 1000; ++i)          f[i] = f[i - 1] + f[i - 1] + f[i - 2] + f[i - 3] ;      int n;      while (~scanf("%d", &n))          puts(f[n].str());      return 0;  }