高精度除法

高精度系列最难

高精度除以低精度

#include<iostream>
#include<string>
using namespace std;
string s;
int a[101], c[101];
long long b;//低精度
long long tmp=0;//存余数
void fanzhuan(string src, int des[])
{
	for (int i = 0; i < src.size(); i++)
	{
		des[i+1] = src[i] - '0';
	}
}

int main()
{
	cin >> s;
	cin >> b;
	int la = s.size();
	fanzhuan(s, a);
	for (int i = 1; i <= la; i++)
	{
		c[i]=(tmp*10+a[i]) / b;
		tmp = (tmp * 10 + a[i]) % b;
	}
	int lc = 1;
	while (c[lc] == 0 && lc < la)
	{
		lc++;
	}
	for (int i = lc; i <= la; i++)
	{
		cout << c[i];
	}

	return 0;
}

利用C语言中的字符串实现,求的是余数

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

bool sub_(char* s1, char* s2)
{
	int n1 = strlen(s1);
	int n2 = strlen(s2);
	if (n1 < n2)return false;
	if (n1 == n2) {
		char* p = s1 + n1 - 1;
		char* q = s2 + n2 - 1;
		while (p >= s1) {
			if (*p > *q) break;
			if (*p < *q)return false;
			p--; q--;
		}
	}
	char* p = s1;
	char* q = s2;

	int cx = 0;//借位标志
	while (*q) {
		int t = *p - *q + cx;
		if (t < 0) {
			cx = -1;
			t += 10;
		}
		else
			cx = 0;
		*p = t + '0';
		p++; q++;
	}
	while (*p) {
		int t = *p - '0' + cx;
		if (t < 0) {
			cx = -1;
			t += 10;
		}
		else
			cx = 0;
		*p = t + '0';
		p++;
	}
	p = s1 + n1 - 1;
	while (p > s1 && *p == '0')p--;
	p[1] = '\0';
	return true;
}



int mod(char* s1, char* s2)
{
	char* p = s1 + strlen(s1) - strlen(s2);
	if (p < s1) p = s1;
	do {
		while (sub_(p , s2));
		p--;
	} while (p >= s1);

	return 0;
}

int main()
{
	char s1[] = "6677834";
	char s2[] = "123456789987654321";
	_strrev(s1);
	_strrev(s2);//反转
	mod(s1, s2);
	_strrev(s1);
	cout << s1 << endl;
	return 0;
}