C++中的高精度运算

c++中的几种高精度运算

A+B problem

P1601 A+B Problem（高精） - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <iostream>
#include <vector>
#include <string>

using namespace std;

vector<int> add(vector<int> &A , vector<int> &B){
	vector<int> c;
	if(A.size()<B.size()) return add(B,A);
	
	int t=0; //进位
	for(int i=0;i<A.size();i++){
		t+=A[i];
		if(i<B.size()) t+=B[i];
		c.push_back(t%10);
		t/=10;
	}
	if(t) c.push_back(1);
	return c;
	
}

int main(){
	string a,b;cin>>a>>b;
	vector<int> A,B;
	
	for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
	for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');
	
	auto c = add(A,B);
	
	for(int i=c.size()-1;i>=0;i--){
		printf("%d",c[i]);
	}
	
	return 0;
}

A-B problem

P2142 高精度减法 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <iostream>
#include <vector>
using namespace std;

//判断是否有A >= B
bool cmp(vector<int> &A,vector<int> &B){
	if(A.size()!=B.size()) return A.size() > B.size();
	for(int i=A.size()-1;i>=0;i++){
		if(A[i]!=B[i]) 
			return A[i]>B[i];
	return true;
}
}

vector<int> sub(vector<int> &A, vector<int> &B){
	vector<int> c;
	int t=0; //进位
	for(int i=0;i<A.size();i++){
		t = A[i] + t;
		if(i<B.size()) t-=B[i];
		c.push_back((t+10)%10);
		if(t<0) t=-1;
		else t=0;
	}
	//除去前导0
	while(c.size()>1&&c.back()==0) c.pop_back();
	return c;
	
}

int main(){
	string a,b; cin>>a>>b;
	
	vector<int> A,B;
	for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
	for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');
	auto c = sub(A,B);
	if(cmp(A,B)) {
		auto c = sub(A,B);
		for(int i=c.size()-1;i>=0;i--) printf("%d",c[i]);
	}
	else{
		printf("-");
		auto c = sub(B,A);
		for(int i=c.size()-1;i>=0;i--) printf("%d",c[i]);
	}
	
	
	return 0;
	
}

A*B problem

P1303 A*B Problem - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 高精度×高精度

#include <iostream>
#include <vector>
using namespace std;
vector<int> mul(vector<int> &A, vector<int> &B) {
	if (A.size() < B.size()) return mul(B, A);
	vector<int> c;
	c.assign(A.size() + B.size(), 0);
	int t = 0; //进位
	for (int i = 0; i < B.size(); i++) {
		for (int j = 0; j < A.size() || t; j++) {
			c[i + j] += A[j] * B[i] + t;
			t = c[i + j] / 10;
			c[i + j] %= 10;
		}
	}
	while (c.size() > 1 && c.back() == 0) c.pop_back();
	return c;
}


int main() {
	string a, b;
	cin >> a >> b;
	vector<int> A, B;
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
	for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
	auto c = mul(A, B);
	for (int i = c.size() - 1; i >= 0; i--) printf("%d", c[i]);
	return 0;
}

高精度×长整数

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> &A, long long b) {
	vector<int> c;
	long long t = 0; //进位
	for (int i = 0; i < A.size() || t; i++) {
		if (i < A.size()) t += A[i] * b;
		c.push_back(t % 10);
		t /= 10;
	}
	while (c.size() > 1 && c.back() == 0) c.pop_back();
	return c;

}
int main() {
	string a;
	cin >> a;
	long long b;
	cin >> b;
	vector<int> A;
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
	auto c = mul(A, b);
	for (int i = c.size() - 1; i >= 0; i--) printf("%d", c[i]);
	return 0;

}

A/B problem

P1480 A/B Problem - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;


vector<int> div(vector<int> &A, long long b, long long &r) {
	vector<int> c;
	r = 0;//余数
	for (int i = A.size() - 1; i >= 0; i--) {
		r = r * 10 + A[i];
		c.push_back(r / b);
		r %= b;
	}
	reverse(c.begin(), c.end());
	while (c.size() > 1 && c.back() == 0) c.pop_back();
	return c;
}
int main() {
	string a;
	cin >> a;
	long long b;
	cin >> b;
	long long r = 0;
	vector<int> A;
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
	auto c = div(A, b, r);
	for (int i = c.size() - 1; i >= 0; i--) printf("%d", c[i]);

	return 0;

}