UVa 465 - Overflow

最新推荐文章于 2017-09-01 19:45:46 发布

原创最新推荐文章于 2017-09-01 19:45:46 发布 · 591 阅读

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本文介绍了解决UVa465-Overflow问题的方法，通过自定义高精度运算类实现大数运算，避免整数溢出，并提供了一个完整的C++实现示例。

题目链接：UVa 465 - Overflow

高精度，注意去了前0再比较，否则是比较不出来或者比较错误的。除了使用高精度，这道题用double也可以做。double范围-1.7*10^(-308) ~ 1.7*10^308。

#include<cstdio>
#include<iostream>
#include <cstring>
using namespace std;

const int maxn = 200;
const int maxm = 2147483647;//(1 << 31) - 1;


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);
    int j;
    for(j = 0;j < len;j++)
    {
        if(num[j] != '0')
            break;
    }
    for(int i = 0; i < len - j; i++)
        s[i] = num[len-i-1] - '0';
    len = len - j;
    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;
}

int main() {
  bign a,b;
  char c;
  char _a[10001];
  char _b[10001];
  while(cin>>_a>>c>>_b)
  {
      a = _a;
      b = _b;
      cout<<_a<<" "<<c<<" "<<_b<<endl;
      if(a > maxm)
        cout<<"first number too big"<<endl;
      if(b > maxm)
        cout<<"second number too big"<<endl;
      if(c == '+' && a + b > maxm)
        cout<<"result too big"<<endl;
      if(c == '*' && a * b > maxm)
        cout<<"result too big"<<endl;
  }
  return 0;
}