高精度模板

高精度加法

题目链接

#include<bits/stdc++.h>
using namespace std;
const int N = 100;

string s1, s2;
int a[N], b[N]; // 也可以使用vector 
int main()
{
	cin >> s1 >> s2;
	reverse (s1.begin(), s1.end());
	reverse (s2.begin(), s2.end());
	int n = s1.size(), m = s2.size();
	for (int i = 0;i < n;i ++) 
		a[i] = s1[i] - '0';
	for (int i = 0;i < m;i ++)
		b[i] = s2[i] - '0';
		
	// 非就地 
	int c[N];
	memset (c, 0, sizeof c);
	for (int i = 0;i < max(n, m);i ++) {
		c[i] += a[i] + b[i];
		c[i+1] = c[i] / 10;
		c[i] %= 10;
	}
	
	int len = max(n, m);
	if (c[len]) len ++;
	for (int i = len-1;i >= 0;i --) cout << c[i]; 
	
	// 就地 1 
	/* 
	int c = 0;
	for (int i = 0;i <= max(n, m);i ++) {
		a[i] += c + b[i];
		c = a[i] / 10;
		a[i] %= 10;
	}
	int len = max(n, m);
	if (a[len]) len ++;
	for (int i = len-1;i >= 0;i --) cout << a[i];
	*/
	
	// 就地 2
	/*
	for (int i = 0;i < max(n, m);i ++) {
		a[i] += b[i];
		a[i+1] += a[i] / 10;
		a[i] %= 10;
	}
	int len = max(n, m);
	if (a[len]) len ++;
	for (int i = len-1;i >= 0;i --) cout << a[i];
	*/
	
	return 0;
}

高精度乘法

高精度乘以高精度

题目链接

不压位

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+10;

string s1, s2;
int a[N], b[N], c[N];
int main()
{
	cin >> s1 >> s2;
	
	reverse (s1.begin(), s1.end());
	reverse (s2.begin(), s2.end());
	
	int n = s1.size(), m = s2.size();
	for (int i = 0;i < n;i ++) 
		a[i] = s1[i] - '0';
	for (int i = 0;i < m;i ++)
		b[i] = s2[i] - '0';
	
	for (int i = 0;i < n;i ++)
		for (int j = 0;j < m;j ++) {
			c[i+j] += a[i] * b[j];
			c[i+j+1] += c[i+j] / 10;
			c[i+j] %= 10;
		}
	int len = n + m - 1;
	while (c[len]) len ++;
	for (int i = len-1;i >= 0;i --) cout << c[i];

	return 0;
}

压位

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+10;

string s1, s2;
int a[N], b[N], c[N];

int main()
{
	cin >> s1 >> s2;
	if (s1 == "0" || s2 == "0") { // !
		cout << 0;
		return 0;
	}
	
	reverse (s1.begin(), s1.end());
	reverse (s2.begin(), s2.end());
	int n = s1.size(), m = s2.size();
	int nn = -1, mm = -1;
	
	for (int i = 0, base = 1;i < n;i ++, base *= 10)
		if (i%4 == 0) a[++nn] = s1[i] - '0', base = 1;
		else a[nn] += base * (s1[i] - '0');
	
	for (int i = 0, base = 1;i < m;i ++, base *= 10)
		if (i%4 == 0) b[++mm] = s2[i] - '0', base = 1;
		else b[mm] += base * (s2[i] - '0');
	
	nn ++, mm ++;

	for (int i = 0;i < nn;i ++)
		for (int j = 0;j < mm;j ++) {
			c[i+j] += a[i] * b[j];
			c[i+j+1] += c[i+j] / 10000; // !
			c[i+j] %= 10000; // !
 		}
	
	int len = nn + mm - 1;
	while (c[len]) len ++;
	cout << c[len-1]; // !
	for (int i = len-2;i >= 0;i --) printf ("%04d", c[i]); // !
	
	
	return 0;
}

高精度乘以低精度

#include<bits/stdc++.h>
using namespace std;
const int N = 100;

string s;
int a[N], x;
int main()
{
	cin >> s >> x;
	
	if (s == "0" || x == 0) {
		cout << 0 << endl;
		return 0;
	}
	
	reverse (s.begin(), s.end());
	int n = s.size();
	
	for (int i = 0;i < n;i ++) 
		a[i] = s[i] - '0';
	
	int c = 0;
	for (int i = 0;i < n;i ++) {
		a[i] = a[i] * x + c;
		c = a[i] / 10;
		a[i] %= 10;
	}

	int len = n;
	while (c) a[len ++] = c%10, c /= 10;
	
	for (int i = len-1;i >= 0;i --) cout << a[i];
	return 0;
}

高精度减法

题目链接

#include<bits/stdc++.h>
using namespace std;
const int N = 10090;

int a[N], b[N], c[N];
string s1, s2, flag = "";

bool change (string s1, string s2) {
	if (s1.size() == s2.size()) return s1 < s2;
	if (s1.size() < s2.size()) return true;
	return false;
}

int main()
{
	cin >> s1 >> s2;
	
	if (s1 == s2) {
		cout << 0 << endl;
		return 0;
	}
	if (change(s1, s2)) flag = "-", swap(s1, s2);
	
	reverse (s1.begin(), s1.end());
	reverse (s2.begin(), s2.end());
	
	int n = s1.size (), m = s2.size ();
	
	for (int i = 0;i < n;i ++) a[i] = s1[i] - '0';
	for (int i = 0;i < m;i ++) b[i] = s2[i] - '0';
	
	for (int i = 0;i < max(n, m);i ++) {
		c[i] += a[i] - b[i];
		if (c[i] < 0) 
			c[i] += 10, c[i+1] --;
	}
	
	int len = max(n, m);
	while (len > 0 && c[len-1] == 0) len --; // len只要 -- 的，一定要加上 >0 的限制，保证全0时也能输出一个0 
	
	cout << flag;
	for (int i = len-1;i >= 0;i --) cout << c[i];

	return 0;
}

高精度除法

有能力的可以去“Acwing-794-高精度除法”提交。

由于没钱，无处提交，不能保证能过。

如果有有钱心善的人愿意帮我提交一下那最好不过了（

高精度除以低精度

#include<bits/stdc++.h>
using namespace std;

const int N = 100;
string s;
int x, m;
int c[N];

int main()
{
	cin >> s >> x;

	int n = s.size (), r = 0;
	for (int i = 0;i < n;i ++) {
		r = r * 10 + (s[i] - '0');
		c[m ++] = r / x;
		r %= x;
	}
	
	int i = 0;
	while (i < m-1 && !c[i]) i ++; // i < m-1：保证全0时也能输出一个0 
	for (;i < m;i ++) cout << c[i];
	
	cout << endl << r << endl;
	
	return 0;
}

高精度除以高精度

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+10;


bool great (string s1, string s2) {
	if (s1.size () == s2.size ()) return s1 >= s2; // >= !!!!
	if (s1.size () > s2.size ()) return true;
	return false;
}

string sub (string s1, string s2) {
	int a[N], b[N], c[N];
	
	memset (a, 0, sizeof a);
	memset (b, 0, sizeof b);
	memset (c, 0, sizeof c);
	
	reverse (s1.begin(), s1.end());
	reverse (s2.begin(), s2.end());
	
	
	int n = s1.size (), m = s2.size ();
	int len = max(n, m);
		
	for (int i = 0;i < n;i ++) a[i] = s1[i] - '0';
	for (int i = 0;i < m;i ++) b[i] = s2[i] - '0';
	
	for (int i = 0;i < len;i ++) {
		c[i] += a[i] - b[i];
		if (c[i] < 0) c[i] += 10, c[i+1] --;
	}
	
	while (c[len-1] == 0) len --;
	string s = "";
	for (int i = len - 1;i >= 0;i --) 
		s += c[i] + '0';
	return s;
}

int main()
{
	string s1, s2;
	
	cin >> s1 >> s2;
	
	string r = "";
	int n = s1.size (), len = 0;
	int c[N];
	memset (c, 0, sizeof c);
	
	for (int i = 0;i < n;i ++) {
		int ans = 0;
		r += s1[i];
		while (great (r, s2)) 
			r = sub (r, s2),
			ans ++;
		c[len ++] = ans;
	} 
	
	int i = 0;
	while (i < len && c[i] == 0) i ++;
	if (i == len) cout << 0;
	else for (;i < len;i ++) cout << c[i];
	
	cout << endl << (r == ""? "0" : r) << endl;

	return 0;
}