探索C/C++大数快(自然数)模板

本文fcbruce个人原创整理。转载请注明出处http://blog.csdn.net/u012965890/article/details/40432511，谢谢。

我们知道在C/C++中int型可处理-2^31~2^31-1(32位及以上编译器)，long long型可处理-2^63~2^63-1的数据,这实际上是非常有限的。在非常多情况下。我们往往会处理范围更大的数据。

Java中有BigInteger类，python中想要多大就有多大（取决于内存），可是C/C++就显得有些乏力，这时候我们会手写大数类。用一个数组记录一个数，来模拟竖式计算。

通常我们会一位一位地储存数据，这样易于实现，逻辑清晰，方便理解，可是一定程度上牺牲了效率，浪费了资源，那么是否能多位存储数据并操作呢。显然是能够的。

我们知道int类型能表示的最大数量级为10^9左右，为了避免乘法溢出，我们最好还是用一个int存储4位数字(10^4)，能够轻易写下例如以下代码（仅含加、减、乘和比較操作）：

/*
 *
 * Author : fcbruce <[e-mail]>
 *
 * Time : Fri 24 Oct 2014 02:43:41 PM CST
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
  #define lld "%I64d"
#else
  #define lld "%lld"
#endif

#define maxm
#define maxn 

using namespace std;

const int maxl = 233;
struct bign
{
  static int width;
  static int mod;
  int len,s[maxl];

  bign()
  {
    memset(s,0,sizeof s);
    len=1;
  }

  bign(int num)
  {
    *this=num;
  }

  bign(long long num)
  {
    *this=num;
  }

  bign(const char *s)
  {
    *this=s;
  }

  bign operator = (int num)
  {
    char s[maxl];
    sprintf(s,"%d",num);
    *this=s;
    return *this;
  }

  bign operator = (long long num)
  {
    char s[maxl];
    sprintf(s,lld,num);
    *this=s;
    return *this;
  }

  bign operator = (const char *s)
  {
    int l=strlen(s);
    len=0;
    for (int i=l-1;i>=0;i-=width,len++)
    {
      (*this).s[len]=0;
      for (int j=max(0,i-width+1);j<=i;j++)
        (*this).s[len]=(*this).s[len]*10+s[j]-'0';
    }

    return *this;
  }

  void str(char *s)
  {
    char format[5];
    sprintf(format,"%%0%dd",width);
    for (int i=len-1,j=0;i>=0;i--,j++)
      sprintf(s+j*width,format,(*this).s[i]);

    int j=0;
    while (s[j]=='0' && s[j+1]!='\0') j++;
    strcpy(s,s+j);
  }

  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%mod;
      g=x/mod;
    }

    return c;
  }

  void clean()
  {
    while (len>1 && s[len-1]==0) 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]/mod;
      c.s[i]%=mod;
    }

    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+=mod;
      }
      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)const
  {
    return !(b>*this);
  }

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

  bool operator == (const  bign &b)const
  {
    if (len!=b.len) return false;
    for (int i=0;i<len;i++)
      if (s[i]!=b.s[i]) return false;
    return true;
  }

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

int bign::width=4;
int bign::mod=10000;

int main()
{
#ifdef FCBRUCE
//  freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE

  int T_T;
  scanf("%d",&T_T);
  bign a,b,c;
  char s1[233],s2[233],s[233];

  while (T_T--)
  {
    scanf("%s %s",s1,s2);
    a=s1;b=s2;
    c=a+b;
    c.str(s);
    printf("%s ",s);

    c=a-b;
    c.str(s);
    printf("%s ",s);

    c=a*b;
    c.str(s);
    printf("%s\n",s);
  }

  return 0;
}

当中void str(char *)函数为将该大数转换成字符串。

随机生成100000组10^82数量级下面的数据并进行对拍，没有发现错误。

long long能表示的数据范围更大，能压很多其它的位数。会不会更快呢？最好还是一次压8位。对以上代码稍加修改就可以

/*
 *
 * Author : fcbruce <[e-mail]>
 *
 * Time : Fri 24 Oct 2014 02:43:41 PM CST
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
  #define lld "%I64d"
#else
  #define lld "%lld"
#endif

#define maxm
#define maxn 

using namespace std;

const int maxl = 233;
struct bign
{
  static int width;
  static long long mod;
  int len;
  long long s[maxl];

  bign()
  {
    memset(s,0,sizeof s);
    len=1;
  }

  bign(int num)
  {
    *this=num;
  }

  bign(long long num)
  {
    *this=num;
  }

  bign(const char *s)
  {
    *this=s;
  }

  bign operator = (int num)
  {
    char s[maxl];
    sprintf(s,"%d",num);
    *this=s;
    return *this;
  }

  bign operator = (long long num)
  {
    char s[maxl];
    sprintf(s,lld,num);
    *this=s;
    return *this;
  }

  bign operator = (const char *s)
  {
    int l=strlen(s);
    len=0;
    for (int i=l-1;i>=0;i-=width,len++)
    {
      (*this).s[len]=0;
      for (int j=max(0,i-width+1);j<=i;j++)
        (*this).s[len]=(*this).s[len]*10+s[j]-'0';
    }

    return *this;
  }

  void str(char *s)
  {
    char format[10];
    sprintf(format,"%%0%d%s",width,lld+1);
    for (int i=len-1,j=0;i>=0;i--,j++)
      sprintf(s+j*width,format,(*this).s[i]);

    int j=0;
    while (s[j]=='0' && s[j+1]!='\0') j++;
    strcpy(s,s+j);
  }

  bign operator + (const bign &b)const
  {
    bign c;
    c.len=0;
    long long g=0ll;
    for (int i=0;g!=0ll || i<max(len,b.len);i++)
    {
      long long x=g;
      if (i<len) x+=s[i];
      if (i<b.len) x+=b.s[i];
      c.s[c.len++]=x%mod;
      g=x/mod;
    }

    return c;
  }

  void clean()
  {
    while (len>1 && s[len-1]==0) 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]/mod;
      c.s[i]%=mod;
    }

    c.clean();
    return c;
  }

  bign operator - (const bign &b)
  {
    bign c;c.len=0;
    long long g=0ll;
    for (int i=0;i<len;i++)
    {
      long long x=s[i]-g;
      if (i<b.len) x-=b.s[i];
      if (x>=0) g=0;
      else
      {
        g=1;
        x+=mod;
      }
      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)const
  {
    return !(b>*this);
  }

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

  bool operator == (const  bign &b)const
  {
    if (len!=b.len) return false;
    for (int i=0;i<len;i++)
      if (s[i]!=b.s[i]) return false;
    return true;
  }

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

int bign::width=8;
long long bign::mod=100000000ll;

int main()
{
#ifdef FCBRUCE
//  freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE

  int T_T;
  scanf("%d",&T_T);
  bign a,b,c;
  char s1[233],s2[233],s[233];

  while (T_T--)
  {
    scanf("%s %s",s1,s2);
    a=s1;b=s2;
    c=a+b;
    c.str(s);
    printf("%s ",s);

    c=a-b;
    c.str(s);
    printf("%s ",s);

    c=a*b;
    c.str(s);
    printf("%s\n",s);
  }

  return 0;
}

改动对拍代码。发现使用压8位的long long 版大数从性能上确实要优于压4位的int版大数，尽管int版偶尔会稍快于long long版，但平均性能上long long版要比int版快20%~60%(包含IO)

数据生成代码：

#
#
# Author : fcbruce <[e-mail]>
#
# Time : Fri 24 Oct 2014 06:33:17 PM CST
#
#

import random

T_T=100000
print T_T
for i in xrange(T_T):
  a=random.randint(0,1237648236422345678987655432349875934875632123131523784682932317237132418743972317);
  b=random.randint(0,12345678987623463824593658235543232123131238746239523172371376382423749824172324317);
  print a+b,a

标程代码：

#
#
# Author : fcbruce <[e-mail]>
#
# Time : Fri 24 Oct 2014 06:38:52 PM CST
#
#

n=input()

for i in xrange(n):
  a,b=map(int,raw_input().split())
  print a+b,a-b,a*b

对拍代码：

#!/bin/bash
#
# Author : fcbruce <[e-mail]>
#
# Time : Fri 24 Oct 2014 07:01:27 PM CST
#
#
while true; do
  python data.py > input
  python std.py < input > std_output

  begin_time_int=$(date "+%s%N")
  ./bign_int < input >bign_int_output
  end_time_int=$(date "+%s%N") 

  begin_time_longlong=$(date "+%s%N")
  ./bign_longlong < input >bign_longlong_output
  end_time_longlong=$(date "+%s%N") 

  use_time_int=$(expr $end_time_int - $begin_time_int)
  use_time_longlong=$(expr $end_time_longlong - $begin_time_longlong)

  echo "bign int"
  diff std_output bign_int_output
  if [ $? -ne 0 ]; then
    echo "Wrong Answer"
    break;
  else
    echo "Accepted"
    echo "run time : " $use_time_int
  fi

  echo

  echo "bign long long"
  diff std_output bign_longlong_output
  if [ $? -ne 0 ]; then
    echo "Wrong Answer"
    break;
  else
    echo "Accepted"
    echo "run time : " $use_time_longlong
  fi

  echo

  test $use_time_longlong -lt $use_time_int
  if [ $? -ne 0 ] ; then
    echo "int faster"
  else
    echo "long long faster"
  fi

  echo
  echo

done

部分測试结果（执行时间单位：十亿分之中的一个秒，測试环境：ubuntu10.04 ,编译开O2优化。gnu++0x。主频：2.26GHz × 2 ）：

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参考资料：

《算法入门经典大赛》—— 刘如家