The University of Zhejiang Pat 3-04. Multiplication and addition of unary polynomial (a solution to the problem)

3-04. Multiplication and addition operations of unary polynomialTime limit MS
Memory Limit 32000 KB
Code length limit 8000 B
Standard of procedure for the award of questions

The product and sum of the two unary polynomials are obtained by the design function respectively.

input Format Description:

Input is divided into 2 lines, each line is given the number of non-0 polynomial, and then the exponential degradation way to enter a polynomial non-0 coefficients and exponents (absolute value is not more than 1000 integers). The numbers are separated by a space.

output Format Description:

The output is divided into 2 rows, and the coefficients and exponents of the product polynomial and the polynomial non-0 are output in exponential degradation mode respectively. The numbers are separated by a space, but cannot have extra spaces at the end.

sample input and output:

Serial number	Input	Output
1	4 3 4-5 2 6 1 -2 0 3 5 -7 4 3 1	24-25 21-10 20-21 8 6-33 5 4-15 3 2-6 1 5 20-4 4-5 2 9 1-2 0
2	2 1 2 1 0 2 1 2-1 0	1 4-1 0 2 2
3	2-1000 0 2 1000 1000-1000 0	-1000000 2000000 1000-1000000 0 0 0
4	0 1 999 1000	0 0 999 1000

The problem is to combine the previous two questions, Pat did not point to innovative consciousness ah, in addition to the code word or code word

#include <iostream> #include <deque> #include <algorithm> using namespace std;
	typedef struct {int XI;
int zhi;

}node;

Deque<node> A, B, C, D;
BOOL Haveflag = false;

bool Firstflag = true;
	void print (Node t) {if (T.xi = = 0) return;
	Haveflag = true;
		if (firstflag) {cout<<t.xi<< "" <<t.zhi;
	Firstflag = false;

} else cout<< "" <<t.xi<< "" <<t.zhi;}
	BOOL CMP (node T1, node T2) {if (T1.zhi > T2.zhi) {return true;
} return false;
	} int main () {int n;
	cin>>n;
		while (n--) {Node T;
		cin>>t.xi>>t.zhi;
	A.push_back (t);
	} cin>>n;
		while (n--) {Node T;
		cin>>t.xi>>t.zhi;
		B.push_back (t);
	D.push_back (t);
	} int i;
		for (i = 0; i < a.size (); ++i) {Node T;
		Int J;
			for (j = 0; J < b.size (); ++j) {t.xi = A[i].xi * B[J].XI;
			T.zhi = A[i].zhi + B[j].zhi; 
			int k;
			BOOL flag = TRUE; for (k = 0; k < c.size (); ++k) {if (c[k].zhi = = T.zhi) {c[k].xi + = T.xi;
				Flag = false;
		}} if (flag) c.push_back (t);
		} bool Flag1 = true;
				for (j = 0; J < d.size (); ++j) {if (D[j].zhi = = A[i].zhi) {d[j].xi + = A[i].xi;
			Flag1 = false;
	}} if (Flag1) D.push_back (A[i]);
	} sort (C.begin (), C.end (), CMP);
	For_each (C.begin (), C.end (), print);
	if (!haveflag) printf ("0 0");
	cout<<endl;
	Firstflag = true;
	Haveflag = false;
	Sort (D.begin (), D.end (), CMP);
	For_each (D.begin (), D.end (), print);
	if (!haveflag) printf ("0 0");
	cout<<endl;
return 0; }