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/article/2059931.html，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 <fcbruce8964@gmail.com>
*
* 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 <fcbruce8964@gmail.com>
*
* 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 <fcbruce8964@gmail.com>
#
# 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 <fcbruce8964@gmail.com>
#
# 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 <fcbruce8964@gmail.com>
#
# 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 ）：



參考资料：

《算法入门经典大赛》—— 刘如家