L2-018 Polynomial A divided by B (analog)

topic link

https://pintia.cn/problem-sets/994805046380707840/problems/994805060372905984

ideas

In fact, it is a polynomial division operation. We try to use BB as much as possible.The maximum item Aof$B\; is\; of\; the\; same\; order\; as\; the\; corresponding\; position\; item\; of\; ,\; and\; then\; a\; subtraction\; operation\; can\; be\; performed.\; For\; details,\; please\; refer\; to\; the\; code.\; Note\; that\; there\; are\; several\; pits\; here:$

• The order of the remainder is less than the order of the quotient: test points 3, 4
• Remove zeros in quotient: test points 1, 4
• The maximum order of B is larger than that of A: test points 2, 3

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

const int N = 2e4 + 10;

double a[N], b[N], q[N];

struct Node {
    
    
	int e;
	double c;
};

void print(vector<Node> &ans) {
    
    
	int l = ans.size();
	cout << l << " ";
	for (int i = 0; i < l; ++i) {
    
    
		cout << ans[i].e << " " << ans[i].c << " \n"[i == l - 1];
	}
	if (!l) cout << "0 0.0" << endl;
}

int n, m;
int main() {
    
    
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cout << fixed << setprecision(1);
	cin >> n;
	double ci;
	int ei, maxe1 = 0, maxe2 = 0;
	for (int i = 1; i <= n; ++i) {
    
    
		cin >> ei >> ci;
		a[ei] = ci;
		maxe1 = max(maxe1, ei);
	}
	cin >> m;
	for (int i = 1; i <= m; ++i) {
    
    
		cin >> ei >> ci;
		b[ei] = ci;
		maxe2 = max(maxe2, ei);
	}
	//除法运算
	for (int i = maxe1; i >= maxe2; --i) {
    
    
		int qe = i - maxe2;
		double qc = a[i] / b[maxe2];
		q[qe] = qc;
		for (int j = maxe2; j >= 0; --j) {
    
    
			int pe = j + qe;
			double pc = qc * b[j];
			a[pe] -= pc;
		}
		a[i] = 0;
	}
	//得到商
	vector<Node> ans;
	for (int i = maxe1 - maxe2; i >= 0; --i)
		if (fabs(q[i]) >= 0.05)
			ans.push_back({
    
    i, q[i]});
	
	print(ans);
	//得到余数
	ans.clear();
	for (int i = maxe1; i >= 0; --i)
		if (fabs(a[i]) >= 0.05)
			ans.push_back({
    
    i, a[i]});
	
	print(ans);
	return 0;
}
/*
2 4 4 3 2
2 5 4 3 2

*/