【算法】高精度加减乘除取模运算 A+B A-B A*B A/B A%B Problem

先上模板

高精度加法

// C = A + B, A >= 0, B >= 0
vector<int> add(vector<int> &a,vector<int> &b){
    //c为答案
    vector<int> c;
    //t为进位
    int t=0;
    for(int i=0;i<a.size()||i<b.size();i++){
        //不超过a的范围添加a[i]
        if(i<a.size())t+=a[i];
        //不超过b的范围添加b[i]
        if(i<b.size())t+=b[i];
        //取当前位的答案
        c.push_back(t%10);
        //是否进位
        t/=10;
    }
    //如果t!=0的话向后添加1
    if(t)c.push_back(1);
    return c;
}

高精度减法

// C = A - B, 满足A >= B, A >= 0, B >= 0
vector<int> sub(vector<int> &A, vector<int> &B)
{
    //答案
    vector<int> C;
    //遍历最大的数
    for (int i = 0, t = 0; i < A.size(); i ++ )
    {
        //t为进位
        t = A[i] - t;
        //不超过B的范围t=A[i]-B[i]-t;
        if (i < B.size()) t -= B[i];
        //合二为一，取当前位的答案
        C.push_back((t + 10) % 10);
        //t<0则t=1
        if (t < 0) t = 1;
       	//t>=0则t=0
        else t = 0;
    }
    //去除前导零
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

高精度比大小（cmp函数）

//高精度比大小
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;
}

高精度乘低精度

// C = A * b, A >= 0, b >= 0
vector<int> mul(vector<int> &A, int b)
{
    //类似于高精度加法
    vector<int> C;
    //t为进位
    int t = 0;
    for (int i = 0; i < A.size() || t; i ++ )
    {
        //不超过A的范围t=t+A[i]*b
        if (i < A.size()) t += A[i] * b;
        //取当前位的答案
        C.push_back(t % 10);
        //进位
        t /= 10;
    }
    //去除前导零
    while (C.size() > 1 && C.back() == 0) C.pop_back();

    return C;
}

高精度乘高精度

vector<int> mul(vector<int> &A, vector<int> &B) {
    vector<int> C(A.size() + B.size() + 7, 0); // 初始化为 0，C的size可以大一点

    for (int i = 0; i < A.size(); i++)
        for (int j = 0; j < B.size(); j++)
            C[i + j] += A[i] * B[j];

    int t = 0;
    for (int i = 0; i < C.size(); i++) { // i = C.size() - 1时 t 一定小于 10
        t += C[i];
        C[i] = t % 10;
        t /= 10;
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back(); // 必须要去前导 0，因为最高位很可能是 0
    return C;
}

高精度除低精度

// A / b = C ... r, A >= 0, b > 0
vector<int> div(vector<int> &A, int b, int &r)//高精度A，低精度b，余数r
{
    vector<int> C;//答案
    r = 0;
    for (int i = A.size() - 1; i >= 0; i -- )
    {
        r = r * 10 + A[i];//补全r>=b
        C.push_back(r / b);//取当前位的答案
        r %= b;//r%b为下一次计算
    }
    reverse(C.begin(), C.end());//倒序为答案
    while (C.size() > 1 && C.back() == 0) C.pop_back();//去除前导零
    return C;
}

高精度除高精度

vector<int> div(vector<int> &A, vector<int> &B, vector<int> &r) {
	vector<int> C;
	if (!cmp(A, B)) {
		C.push_back(0);
		r.assign(A.begin(), A.end());
		return C;
	}
	int j = B.size();
	r.assign(A.end() - j, A.end());
	while (j <= A.size()) {
		int k = 0;
		while (cmp(r, B)) {
			r = sub(r, B);
			k ++;
		}
		C.push_back(k);
		if (j < A.size())
			r.insert(r.begin(), A[A.size() - j - 1]);
		if (r.size() > 1 && r.back() == 0)
			r.pop_back();
		j++;
	}
	reverse(C.begin(), C.end());
	while (C.size() > 1 && C.back() == 0)
		C.pop_back();
	return C;
}

再上题目

P1932 A+B A-B A*B A/B A%B Problem

最后AC代码

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

using namespace std;

//高精度比大小
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;
}

//高精度加法
// C = A + B, A >= 0, B >= 0
vector<int> add(vector<int> &a, vector<int> &b) {
	//c为答案
	vector<int> c;
	//t为进位
	int t = 0;
	for (int i = 0; i < a.size() || i < b.size(); i++) {
		//不超过a的范围添加a[i]
		if (i < a.size())
			t += a[i];
		//不超过b的范围添加b[i]
		if (i < b.size())
			t += b[i];
		//取当前位的答案
		c.push_back(t % 10);
		//是否进位
		t /= 10;
	}
	//如果t!=0的话向后添加1
	if (t)
		c.push_back(1);
	return c;
}

//高精度减法
// C = A - B, 满足A >= B, A >= 0, B >= 0
vector<int> sub(vector<int> &A, vector<int> &B) {
	//答案
	vector<int> C;
	//遍历最大的数
	for (int i = 0, t = 0; i < A.size(); i ++ ) {
		//t为进位
		t = A[i] - t;
		//不超过B的范围t=A[i]-B[i]-t;
		if (i < B.size())
			t -= B[i];
		//合二为一，取当前位的答案
		C.push_back((t + 10) % 10);
		//t<0则t=1
		if (t < 0)
			t = 1;
		//t>=0则t=0
		else
			t = 0;
	}
	//去除前导零
	while (C.size() > 1 && C.back() == 0)
		C.pop_back();
	return C;
}

//高精度除法
vector<int> mul(vector<int> &A, vector<int> &B) {
	vector<int> C(A.size() + B.size() + 7, 0); // 初始化为 0，C的size可以大一点

	for (int i = 0; i < A.size(); i++)
		for (int j = 0; j < B.size(); j++)
			C[i + j] += A[i] * B[j];

	int t = 0;
	for (int i = 0; i < C.size(); i++) { // i = C.size() - 1时 t 一定小于 10
		t += C[i];
		C[i] = t % 10;
		t /= 10;
	}

	while (C.size() > 1 && C.back() == 0)
		C.pop_back(); // 必须要去前导 0，因为最高位很可能是 0
	return C;
}

vector<int> div(vector<int> &A, vector<int> &B, vector<int> &r) {
	vector<int> C;
	if (!cmp(A, B)) {
		C.push_back(0);
		r.assign(A.begin(), A.end());
		return C;
	}
	int j = B.size();
	r.assign(A.end() - j, A.end());
	while (j <= A.size()) {
		int k = 0;
		while (cmp(r, B)) {
			r = sub(r, B);
			k ++;
		}
		C.push_back(k);
		if (j < A.size())
			r.insert(r.begin(), A[A.size() - j - 1]);
		if (r.size() > 1 && r.back() == 0)
			r.pop_back();
		j++;
	}
	reverse(C.begin(), C.end());
	while (C.size() > 1 && C.back() == 0)
		C.pop_back();
	return C;
}

int main() {
	string a, b;
	vector<int> A, B, r;
	//读入高精度
	cin >> 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');
	//高精度加法输出
	auto C1 = add(A, B);
	for (int i = C1.size() - 1; i >= 0; i -- )
		printf("%d", C1[i]);
	cout << endl;
	//高精度减法输出
	if (cmp(A, B)) {
		auto C2 = sub(A, B);
		for (int i = C2.size() - 1; i >= 0; i -- )
			printf("%d", C2[i]);
		cout << endl;
	} else {
		auto C3 = sub(B, A);
		printf("-");
		for (int i = C3.size() - 1; i >= 0; i -- )
			printf("%d", C3[i]);
		cout << endl;
	}
	//高精度乘法输出
	auto C4 = mul(A, B);
	for (int i = C4.size() - 1; i >= 0; i--)
		printf("%d", C4[i]);
	cout << endl;
	//高精度除法输出
	auto C5 = div(A, B, r);
	for (int i = C5.size() - 1; i >= 0; i--)
		printf("%d", C5[i]);
	cout << endl;
	for (int i = r.size() - 1; i >= 0; i--)
		printf("%d", r[i]);
	cout << endl;
	return 0;
}