大数乘法。大数运算。“无限大数字”乘法。大数乘法两种方法对比

最近在看笔试题，得知大数运算是个经常考的题目。所以有兴趣试了试。

一开始按照笔算方法自己写了个，但是时间复杂度是o(n3)。

参考了网上的算法之后，修改了自己的算法，时间复杂度变成o(n2)。

下面的测试结果中，两个2000位的数字（阿拉伯数字的位数）相乘，耗时90多毫秒。

200位，1毫秒。可以看到，复杂度的确是N的平方级别。

自己写的笨办法，每次累加之后都要判断是否有进位。但是安全。

网上有个高效的算法，使用int存储临时结果，不用每次累加后都判断一次进位。等

所有的累加都完成之后再判断，所以时间复杂度降低了一个数量级。

但是这样也有个坏处，就是int早晚有溢出的时候，当整数的位数足够多，

也就是达到了2的29、30、31、32次方位（当然这种情况基本没可能发生），这种方法的运算结果就是错误的了。

下面是算法对应的函数。

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char* BigIntMultiply ( const char * const a, const char * const b , char* const lresult)

{

int i,j;

int la = strlen(a);

int lb = strlen(b);

int rlen = la+lb;

int* r = (int*)calloc( rlen, sizeof(int) );

for(i = 0;i < lb; i++)

for(j = 0; j < la; j++)

r[rlen - 1 - i - j] += (b[lb - i - 1] - '0') * (a[la - j - 1] - '0');

//then is there carry on current number

for(j = rlen - 1; j >= 1; j--)

if(r[j] > 9)

{

r[j-1] += r[j] / 10;

r[j] %= 10;

}

//find first none_zero_index

for(i = 0; 0 == r[i]; i++){}

//mem cpy

for(j=0; i< rlen; i++,j++)

lresult[j] = r[i]+'0';

lresult[j]='\0';

free(r);

return lresult;

}

下面的代码在Visual Studio 2008里面编译运行，没有问题。 Linux上没有 SYSTEMTIME，

没有atoi，itoa，GetLocalTime。所以要在Linux运行，得相应的修改一下。

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#include <iostream>

#include <assert.h>

#include <stdio.h>

#include <time.h>

#include <windows.h>

#include <malloc.h>

using namespace std;

const int MAX = 2001;

char num1[MAX];

char num2[MAX];

char result[2*MAX];

void SafeGetNumFromStr ( char* num, char* str);

char* BigIntMultiply ( const char * const a, const char * const b , char* const lresult);

void multiply( const char *a, const char *b);

int main(int argc, char *argv[])

{

//test speed

cout<<"\n\nspeed test... Number of digits is : "<<MAX-1<<"\n";

int i;

const int TEST_TIME = 20;

srand((unsigned)time(NULL));

for(i = 0;i<MAX;i++)

{

num1[i] = 0;

num2[i] = 0;

}

//create data with random

for(i = 0; i< MAX - 1; i++)

{

num1[i] = rand()%10 + '0';

num2[i] = rand()%10 + '0';

}

SYSTEMTIME wtm;

GetLocalTime(&wtm);

long long time_start = wtm.wMilliseconds + wtm.wSecond * 1000;

cout<<num1<<endl;

cout<<"*\n";

cout<<num2<<endl;

for(i = 0; i<TEST_TIME; i++)

{

BigIntMultiply(num1,num2,result);

}

GetLocalTime(&wtm);

cout<<"Result is:\n";

cout<<result<<endl;

double tmv = (double)(wtm.wMilliseconds + wtm.wSecond * 1000 - time_start);

cout<<"Test Over. "<<TEST_TIME<<" loops use time: "<<tmv<<" ms\n";

cout<<" Each One Time Use: "<<tmv/(double)TEST_TIME<<" ms\n\n\n";

//test validation

cout<<"Validation work...\n";

long long testNum1;

long long testNum2;

int testI;

for(testNum1 = 0;testNum1<1000000000000000;testNum1 = (testNum1+1)*181+1)

for(testI= 0;testI<200; testI++)

{

char a[2*MAX];

char b[2*MAX];

testNum2 = (testNum1+testI)<0?0:testNum1+testI;

for(i = 0;i<MAX;i++)

{

num1[i] = 0;

num2[i] = 0;

}

itoa(testNum1,a,10);

itoa(testNum2,b,10);

SafeGetNumFromStr(num1,a);

SafeGetNumFromStr(num2,b);

BigIntMultiply(num1,num2,result);

if(8 == testNum2%10)

if(testNum1*testNum2 == atoi(result))

cout<<testNum1<<" * "<<testNum2<<" == "<<testNum1*testNum2<<" Correct!\n";

else

cout<<testNum1<<" * "<<testNum2<<" Result:"<<result<<"\n";

}

//free test

cout<<"Now ..... Free Test!\n";

while(1)

{

char a[2*MAX];

char b[2*MAX];

cout<<"\n\ninput long integer for A"<<endl;

cin>>a;

cout<<"input long integer for B"<<endl;

cin>>b;

//get data

SafeGetNumFromStr(num1,a);

SafeGetNumFromStr(num2,b);

cout<<endl<<endl;

cout<<num1;

cout<<" * ";

cout<<num2;

cout<<endl;

BigIntMultiply(num1,num2,result);

cout<<"Result is:"<<endl;

cout<<result;

}

system("pause");

return 0;

}

void SafeGetNumFromStr( char* num, char* str)

{

memset(num,0,sizeof(num[0])*MAX);

int i;

int index = 0;

for(i=0;i<2*MAX && index < MAX;i++)

{

if(str[i] <= '9' && str[i] >='0')

num[index++] = str[i];

if('\0'==str[i])

break;

}

assert( 0 != index );

}

char* BigIntMultiply ( const char * const a, const char * const b , char* const lresult)

{

int i,j;

int la = strlen(a);

int lb = strlen(b);

int rlen = la+lb;

int* r = (int*)calloc( rlen, sizeof(int) );

for(i = 0;i < lb; i++)

for(j = 0; j < la; j++)

r[rlen - 1 - i - j] += (b[lb - i - 1] - '0') * (a[la - j - 1] - '0');

//then is there carry on current number

for(j = rlen - 1; j >= 1; j--)

if(r[j] > 9)

{

r[j-1] += r[j] / 10;

r[j] %= 10;

}

//find first none_zero_index

for(i = 0; 0 == r[i]; i++){}

//mem cpy

for(j=0; i< rlen; i++,j++)

lresult[j] = r[i]+'0';

lresult[j]='\0';

free(r);

return lresult;

}