【vijos1033】【数值/数论】【高精度乘法】整数分解(版本2)

描述 Description
	整数分解(版本2)  一个正整数可以分解成若干个自然数之和。请你编一个程序，对于给出的一个正整数n(1<=n<=1500)，求出满足要求的分解方案，并使这些自然数的乘积m达到最大。  例如n=10，则可以分解为2+2+3+3，乘积m=2*2*3*3=36

输入格式 Input Format
	一个正整数n

输出格式 Output Format
	输出分解的自然数的最大乘积m

	样例输入 Sample Input

10

样例输出 Sample Output

36

这道题考察了数学和高精度 由于分析很多所以就用图片

就像上面那样…

可以得到N关于M的表达式

然后特判一下M＝1的情况就AC了

很不错的一道题

代码如下

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 2000;
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 = (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] - '0';
    return *this;
  }

  string str() const {
    string res = "";
    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
    if(res == "") res = "0";
    return res;
  }

  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;
  }

  void clean() {
    while(len > 1 && !s[len-1]) len--;
  }

  bign operator * (const bign& b) {
    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;
  }

  bign operator - (const bign& b) {
    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;
  }

  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) {
    return !(b > *this);
  }

  bool operator == (const bign& b) {
    return !(b < *this) && !(*this < b);
  }

  bign operator += (const bign& b) {
    *this = *this + b;
    return *this;
  }
};

istream& operator >> (istream &in, bign& x) {
  string s;
  in >> s;
  x = s.c_str();
  return in;
}

ostream& operator << (ostream &out, const bign& x) {
  out << x.str();
  return out;
}

void init_file()
{
	freopen("divide.in", "r", stdin);
	freopen("divide.out", "w", stdout);
}

int N;
bign ans = 0;

void read_data()
{
	scanf("%d", &N);
	if(N == 1)
	{
		printf("1");
		exit(0);
	}
}

void work()
{
	if (N % 3 == 1)
	{
		ans = 2 * 2;
		for(int i = 1; i <= (N - 4) / 3; i++)
		{
			ans = ans * 3;
		}
		cout << ans;
		exit(0);
	}
	if (N % 3 == 2)
	{
		ans = 2;
		for(int i = 1; i <= (N - 2) / 3; i++)
		{
			ans = ans * 3;
		}
		cout << ans;
		exit(0);
	}
	if (N % 3 == 0)
	{
          ans = 1;
          for(int i = 1; i <= N / 3; i++)
          {
                  ans = ans * 3;
          }
          cout << ans;
          exit(0);
    }
}

int main() 
{
	init_file();
	read_data();
	work();
	return 0;
}