板子

高精度加法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int a[255], b[255], c[255];

int main()
{
	string s1, s2;
	int la, lb, len;
	/*memset(a, 0, sizeof(a));    
	memset(b, 0, sizeof(b));
	memset(c, 0, sizeof(c));*/
	
	cin >> s1 >> s2;
	la = s1.size();
	lb = s2.size();
	for (int i = la - 1; i >= 0; i--) a[la - i - 1] = s1[i] - '0';
	for (int i = lb - 1; i >= 0; i--) b[lb - i - 1] = s2[i] - '0';
	
	if(la > lb) len = la;
	else len = lb;
	
	for (int i = 0; i < len; i++) c[i] = a[i] + b[i];
	
	for (int i = 0; i < len; i++)
	{
		c[i + 1] += c[i] / 10;
		c[i] %= 10;
	}
	
	if (c[len] > 0) len++;
	
	/*for(int i = 0;i < len;i++) 
    std :: cout << c[i] << ' ';
    cout << endl;
	*/
	for (int i = len - 1; i >= 0; i--) cout << c[i];
	
	return 0;
}

高精度减法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int a[10086], b[10086], c[10086];

int main()
{
	string s1, s2, s;
	char fh;
	int la, lb, k;
	
	cin >> s1 >> s2;
	if (s1 == s2) 
	{
		cout << "0" << endl;
	    return 0;
	}
	
	la = s1.size();
	lb = s2.size();
	
	if (la < lb || ((la == lb) && (s1 < s2)))
	{
		fh = '-';
	    s = s1;
	    s1 = s2;
	    s2 = s;
	}
	
	
	la = s1.size();
	lb = s2.size();//la绝对值大 
	
	for (int i = la - 1; i >= 0; i--) a[la - i - 1] = s1[i] - '0';
	for (int i = lb - 1; i >= 0; i--) b[lb - i - 1] = s2[i] - '0';
	k = la - 1;
	
	for (int t = 0; t < la; t++)
	{
		if (a[t] < b[t])
		{
			a[t + 1] -= 1;
			a[t] += 10;
		}
		a[t] -= b[t];
	}
	
	while (a[k] == 0) k--;
	
	/*for(int i = 0;i < len;i++) 
    std :: cout << c[i] << ' ';
    cout << endl;
	*/
	
	if (fh == '-') cout << fh;
	for (int i = k; i >= 0; i--) cout << a[i];
	
	return 0;
}

高精度乘法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int a[2001], b[2001], c[4001];

int main()
{
	string s1, s2;
	int la, lb, len;
	/*memset(a, 0, sizeof(a));    
	memset(b, 0, sizeof(b));
	memset(c, 0, sizeof(c));*/
	
	cin >> s1 >> s2;
	la = s1.size();
	lb = s2.size();
	len = la + lb;
	for (int i = la - 1; i >= 0; i--) a[la - i] = s1[i] - '0';
	for (int i = lb - 1; i >= 0; i--) b[lb - i] = s2[i] - '0';//从1开始 
	
	for (int i = 1; i <= la; i++)
	{
		for (int j = 1; j <= lb; j++)
		{
			c[i + j - 1] += a[i] * b[j]; //模拟竖式计算 
		}
	}
	
	for (int i = 1; i <= len; i++)
	{
		if (c[i] >= 10)
		{
			c[i + 1] += c[i] / 10;
			c[i] %= 10;//进位处理 
		}
	}
	int f = 0;
	for (int i = len; i >= 1; i--)
	{
		if (c[i])
		{
			len = i;
			break;
		}
		if (i == 1) f = 1;
	}
	if (f == 1) cout << 0;
	else for (int i = len; i >= 1; i--) cout << c[i];
	
	return 0;
}

高精度除法

#include <iostream>  
#include <cstring>  
#include <cstdio>  

using namespace std;

const int N = 100001;
int a[N], b[N], res[N * 2];
char x[N], y[N];
int la, lb;

void init()
{
	la = strlen(x);
	lb = strlen(y);
	memset(a,0,sizeof(a));  
    memset(b,0,sizeof(b)); 
    memset(res,0,sizeof(res)); 
    
    for (int i = la - 1; i >= 0; i--) a[la - i - 1] = x[i] - '0';
    for (int i = lb - 1; i >= 0; i--) b[lb - i - 1] = y[i] - '0';
}

int substraction(int *p1,int *p2,int len1,int len2)
//计算长度为len1的大整数减去长度为len2的大整数的结果的长度 
{
	//减的结果放在数组p1中，不够返回-1,正好返回0 
	if (len1 < len2) return -1;
	bool flag = 0;
	if (len1 == len2)
	{
		for (int i = len1 - 1; i >= 0; i--)
		{
			if (p1[i] > p2[i]) flag = 1;
			else if (p1[i] < p2[i])
			{
				if (!flag) return -1;
			}
		}
	}
	
	for (int i = 0; i < len1; i++)
	{
		p1[i] -= p2[i];
		if (p1[i] < 0)
		{
			p1[i] += 10;
			p1[i + 1]--;
		}
	}
	for (int i = len1 - 1; i >= 0; i--)
	{
		if (p1[i]) return i + 1;
	}
	return 0;
}

void output()
{
	for (int i = 0; i < N; i++)
	{
		if (res[i] >= 10)//进位 
		{
			res[i + 1] += res[i] / 10;
			res[i] %= 10;
		}
	}
	
	bool flag = 0;
	for (int i = N - 1; i >= 0; i--)
	{
		if (flag) cout << res[i];
		else if (res[i])
		{
			cout << res[i];
			flag = 1;
		}
	}
	
	if (!flag) cout << 0;
	
	cout << endl;
}

void divv()
{
	init();
	
	if (la < lb)
	{
		cout << 0 << endl;
		return;
	}
	
	la = substraction(a, b, la, lb);
	if (la < 0)
	{
		cout << 0 << endl;
		return;
	}
	else if (la == 0)
	{
		cout << 1 << endl;
		return;
	}
	
	res[0]++;//减掉一次了，商+1
	int k = la - lb;
	if (k < 0)//减一次后不能再减了 
	{
		output();
		return;
	}
	else if (k > 0)//将数组b乘以10的某次幂，使得其长度与数组a相同
	{
		for (int i = la - 1; i >= 0; i--)
		{
			if (i >= k) b[i] = b[i - k];
			else b[i] = 0;
		}
	}
	
	lb = la;
	for (int i = 0; i <= k; i++)//先减去若干个b*(10^k)，不够减了再减去若干个b*(10^(k-1))
	{
		int temp;
		while ((temp = substraction(a, b + i, la, lb - i)) >= 0)//一直减到不够减为止
		{
			la = temp;
			res[k - i]++; //每成功减一次，则商的相应位+1
		}
	}
	output();
}

int main()
{
	scanf("%s %s", x, y); 
	divv();
	//init();
	//for (int i = 0; i < la; i++) cout << a[i];
	return 0;
}