俺求个大数除法。。。。。

[elmnd]

１。　带注解
２。　用数组或者链表存放的大数
３。　最好输入2个大数的部分封装一下，　而不是已经赋初值。
４。　不要混乱代码。。。

[xiaoyisnail]

先mark

[帅得不敢出门]

up

[太乙]

http://d.download.csdn.net/down/403510/hqin6

[keven1868]

-_-!

[太乙]

http://d.download.csdn.net/down/403510/hqin6

俺以前写的，很糙！将就看吧！嘿嘿~~

[太乙]

ps：满足lz的所有要求~

[baihacker]

#include <stdio.h>

typedef struct BigNum_
{
	int len;
	int data[1005];
	int sign;
} BigNum;

void Zero(BigNum* num)
{
	int i = 0;
	int* data = num->data;
	for (i = 0; i < 1000; ++i) data[i] = 0;
	num->sign = 1;
}

void Rev(char* a)
{
	char t;
	char* b = a;
	while (*b++);
	b-=2;
	while (a<b) t = *a, *a=*b, *b = t, ++a, --b;
}

void ToBigNum(BigNum* num, char* str)
{
	int len = 0;
	int index = 0;
	int token;
	int base;
	int* data = num->data;
	if (*str=='-') num->sign = -1, str++;
	else num->sign = 1;
	Rev(str);
	while (*str)
	{
		token = 0;
		base = 1;
		while (*str&&index++<4) token = token+base*(*str++-'0'), base*=10;
		index = 0;
		data[len++] = token;
	}
	if (len == 0) data[0]=0, num->len = 1;
	else num->len=len;
}



void Print(BigNum* it)
{
	int curr = it->len-1;
	if (it->sign == -1) printf("-");
	if (curr < 0) printf("0");
	else
	{
	if (curr>=0) printf("%d", it->data[curr--]);
	while (curr>=0) printf("%04d", it->data[curr--]);
	}
	puts("");
}

char str_buff[2][1000];
BigNum a, b, t, result;
int temp[1000];
int sub(int* buff, int hi, int lo, int* buff1, int hi1, int lo1, int it)
{
	int i, j, k, l = hi1+1;
	int inc = 0;
	int hi2;
	#ifdef TEST
	printf("subtest\t%d\t%d\t%d\t%d\n", hi, lo, hi1, lo1);
	#endif
	for (i = lo1; i <= hi1 +5; ++i) temp[i] = 0;
	for (i = lo1; i <= hi1; ++i)
	{
		int k = it * buff1[i] + inc;
		temp[i] = k % 10000;
		inc = k / 10000;
	}
	if (inc) temp[l++] = inc;
	hi2 = l - 1;
	while (buff[hi] == 0 && hi >= lo) -- hi;
	if (l > hi - lo + 1) return 0;
	if (hi - lo == hi2 - lo1)
	for (i = hi, j = hi2; i >= lo; --i, --j)
		{
			#ifdef TEST
			printf("cmp\t%d\t%d\n", buff[i], temp[j]);
			#endif
			if(buff[i] < temp[j]) return 0;
			if (buff[i] > temp[j]) break;
		}
	for (i = lo1, j = lo; i <= hi2; ++i,++j)
	{
		#ifdef TEST
		printf("sub\t%d\t%d\n", buff[j], temp[i]);
		#endif 
		if (buff[j] >= temp[i]) buff[j]-= temp[i];
		else
		{
			while (buff[j] < temp[i]) buff[j] += 10000, --buff[j+1];
			buff[j] -= temp[i];
		}
	}
	return 1;
}
void Div(BigNum* result, BigNum* a, BigNum* b)
{
	int la = a->len;
	int lb = b->len;
	int lr = la - lb + 1;
	int curr = lr - 1;
	int hia = la - 1;
	int loa = curr;
	int hib = lb - 1;
	int lob = 0;
	int* abuff = a->data;
	int* bbuff = b->data;
	int* rbuff = result->data;
	/*确定curr位上的商*/
	for (;curr>=0;--curr)
	{
		int test;/*试商*/
		#ifdef TEST
		printf("%d\t%d\n%d\t%d\n", hia, loa, hib, lob);
		#endif

		/*除数至少有两位*/
		if (lb > 1)
		{	/*被除数中所要考虑的位数大于除数的位数*/
			if (hia-loa>hib-lob)
			{
				test = (abuff[hia]*10000 + abuff[hia-1]) / bbuff[hib];
			}
			else/*被除数中所要考虑的等于大于除数的位数*/
			{
				/*printf("test\t%d\t%d\n", abuff[hia]*10000 + abuff[hia-1], bbuff[hib]*10000 + bbuff[hib-1]);*/
				test = (abuff[hia]*10000 + abuff[hia-1]) / (bbuff[hib]*10000 + bbuff[hib-1]);
			}
		}
		else
		{
			if (hia-loa+1>1)
			{
				test = (abuff[hia]*10000 + abuff[hia-1]) / bbuff[hib];
			}
			else
			{
				/*printf("test\t%d\t%d\n", abuff[hia]*10000 + abuff[hia-1], bbuff[hib]*10000 + bbuff[hib-1]);*/
				test = abuff[hia]/bbuff[hib];
			}
		}

		#ifdef TEST
		printf("%d\n", test);
		#endif
		for( ; test > 0; --test)
		{
			if(sub(abuff, hia, loa, bbuff,  hib, lob, test))
			{
				
				break;
			}
		}
		if (abuff[hia] == 0) -- hia;
		--loa;
		#ifdef TEST
		printf("use%d\n", test);
		#endif
		rbuff[curr] = test;
	}
	
	/*首位为零*/
	curr = lr - 1;
	while (result->data[curr] == 0) --curr;
	result->len = curr + 1;
	result->sign = a->sign * b->sign;
}
int main(void)
{
	
	while (scanf("%s%s", str_buff[0], str_buff[1])==2) 
	{
		ToBigNum(&a, str_buff[0]);
		ToBigNum(&b, str_buff[1]);
		Zero(&result);
		Div(&result, &a, &b);
		Print(&result);
	}
    return 0;
}

[baihacker]

int la = a->len;
int lb = b->len;
int lr = la - lb + 1;
int curr = lr - 1;
int hia = la - 1;
int loa = curr;
int hib = lb - 1;
int lob = 0;
int* abuff = a->data;
int* bbuff = b->data;
int* rbuff = result->data;

l前缀表示len,也就是说a 是被除数,b是除数,r是结果(result)

loa 到 hia的部分除以lob 到hib的部分 商放在curr处

/*除数至少有两位*/
此注释以下在进行试商,分了一些情况在讨论,代码可以更简化的.


int sub(int* buff, int hi, int lo, int* buff1, int hi1, int lo1, int it)

这个函数表示

buff从lo到hi的位置表示的数 - (buff1从lo1到hi1的位置表示的数) * it

如果能减尽的话,执行减法,并返回1.
否则返回0.

for (i = lo1; i <= hi1 +5; ++i) temp[i] = 0;
for (i = lo1; i <= hi1; ++i)
{
int k = it * buff1[i] + inc;
temp[i] = k % 10000;
inc = k / 10000;
}
if (inc) temp[l++] = inc;
这是在计算减数

hi2 = l - 1;
while (buff[hi] == 0 && hi >= lo) -- hi;
if (l > hi - lo + 1) return 0;
if (hi - lo == hi2 - lo1)
for (i = hi, j = hi2; i >= lo; --i, --j)
{
#ifdef TEST
printf("cmp\t%d\t%d\n", buff[i], temp[j]);
#endif
if(buff[i] < temp[j]) return 0;
if (buff[i] > temp[j]) break;
}
这是在判断被减数是不是大于减数

for (i = lo1, j = lo; i <= hi2; ++i,++j)
{
#ifdef TEST
printf("sub\t%d\t%d\n", buff[j], temp[i]);
#endif
if (buff[j] >= temp[i]) buff[j]-= temp[i];
else
{
while (buff[j] < temp[i]) buff[j] += 10000, --buff[j+1];
buff[j] -= temp[i];
}
}
执行减法

[珍惜生命远离CPP]

http://blog.csdn.net/akirya/archive/2008/02/04/2083998.aspx

[chen_de_sheng]

http://enjoy-math.equn.com/Math/Mathematical/Inverse.htm
先用牛顿法求倒数，再做乘法。

[chen_de_sheng]

http://enjoy-math.equn.com/Programs/BigInt.txt

[chen_de_sheng]

http://enjoy-math.equn.com/index.htm - -｜｜｜

[「已注销」]

mark

[viprs]

mark!

[viprs]

mark!

[星羽]

http://www.cppblog.com/elva/archive/2007/08/15/30070.html

[就呆在云上]

ding!

[neeestth]

[guowei6229067]

好想看