用c++实现合并两个一元多项式并求一介导

#include

using namespace std;
#define MAXSIZE 1000

int main() {
	int a[MAXSIZE][2];
	int b[MAXSIZE][2];
	int num1,num2;
	cin >> num1 ;
	for (int i = 0; i < num1; i++) {
		cin >> a[i][0] >> a[i][1];
	}
	cin >> num2;
	for (int i = 0; i < num2; i++) {
		cin >> b[i][0] >> b[i][1];
	}
	if (num1 == 0 && num2 == 0) {
		cout << "0 0"; return 0;
	}
	else if (num1 == 0) {
		for (int i = 0; i < num2; i++) {
			if (i == num2 - 1) 
				cout << b[i][0] << " " << b[i][1];
			else cout << b[i][0] << " " << b[i][1] << " ";
		}
		return 0;
	}
	else if (num2 == 0) {
		for (int i = 0; i < num1; i++) {
			if (i == num1 - 1)
				cout << a[i][0] << " " << a[i][1];
			else cout << a[i][0] << " " << a[i][1] << " ";
		}
		return 0;
	}
	int c[MAXSIZE][2];
	//两个多项式求一介导再相加，存到c中
	int size = 0;
	int i = 0,j = 0;
	while (i < num1 && j < num2) {
		if ( a[i][1] - 1 > b[j][1] - 1) {
			c[size][0] = a[i][0] * a[i][1];
			c[size][1] = a[i][1] - 1;
			size++;
			i++;
		}
		else if ( a[i][1] - 1 < b[j][1] - 1) {
			c[size][0] = b[j][0] * b[j][1];
			c[size][1] = b[j][1] - 1;
			size++;
			j++;
		}
		else  {
			c[size][0] = b[j][0] * b[j][1]+ a[i][0] * a[i][1];
			c[size][1] = b[j][1] - 1;
			size++;
			j++; i++;
		}
	}
	if (a[i][1]) {
		c[size][0] = a[i][0] * a[i][1];
		c[size][1] = a[i][1] - 1;
		size++;
		i++;
	}
	else if (b[j][1]) {
		c[size][0] = b[j][0] * b[j][1];
		c[size][1] = b[j][1] - 1;
		size++;
		j++;
	}
	size--;
	for (int i = 0; i < size; i++) {
		if (i == size - 1)
			cout << c[i][0] << " " << c[i][1];
		else cout << c[i][0] << " " << c[i][1] << " ";
	}
	return 0;
}