pat甲级1009:Product of Polynomials

This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < …< N2 < N1 <=1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

翻译:
这一次,你要做的是找到A和B这两个多项式的乘积.

输入格式:

每个输入文件都包含一个测试用例。 每一行占用2行，每一行包含多项式的信息：其中K是多项式中非零项的个数，Ni和aNi（i = 1,2，…，N）。 …，K）分别是指数和系数。 其中1 <= K <= 10, 0 <= NK < …< N2 < N1 <=1000.

输出格式:
对于每个测试用例，你应该在一行中输出A和B的乘积，格式与输入相同。 请注意，每行末尾不得有额外的空格。 精确到小数点后1位。

思路:
使用链表来表示多项式A,B

步骤:

1. 使用两次循环,将B的每一项与A的每一项相乘,得到链表P
2. 多项式降幂排列: 遍历P的指数成员,将其存在一个数组中,然后对这个数组排序(降序),再依次在链表中查找数组中每一个元素,把每一个节点复制到新的链表中.
3. 合并同类项: 在上一步中已经把多项式降幂排列了,所以如果某个节点的指数成员和下一个节点的指数成员相等,就可以把它们合并成一项.
4. 删除系数为零的项:遍历链表,如果某个节点的系数成员为零,则删除这一项.

代码(我的环境为ms vs2017):

#include "stdafx.h"#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <time.h>#include <iomanip>#include <math.h>using namespace std;struct LinkedList{    int Exp;//指数    double Coe;//系数    LinkedList* Next;};typedef LinkedList* List;int cmp(const void* a, const void* b){    return *(int*)a - *(int*)b;}class Stack{private:    int NodeNumbers;    List L;public:    Stack()    {        NodeNumbers = 0;        L = new LinkedList;        L->Coe = L->Exp = 0;        L->Next = NULL;    }    bool IsEmpty()    {        return L->Next == NULL;    }    void Push(int Exp, double Coe)    {        List Tmp = new LinkedList;        Tmp->Coe = Coe;        Tmp->Exp = Exp;        Tmp->Next = L->Next;        L->Next = Tmp;        NodeNumbers++;    }    void Pop()    {        if (IsEmpty())        {            cout << "Empty stack!" << endl;            return;        }        List Tmp = L->Next;        L->Next = Tmp->Next;        delete Tmp;        NodeNumbers--;    }    ~Stack()    {        List P = L->Next;        while (P != NULL)        {            L->Next = P->Next;            delete P;            P = L->Next;        }        delete L;    }    void Show()    {        if (NodeNumbers == 0)        {            cout << "Empty stack!" << endl;            return;        }        List Tmp = L->Next;        cout << NodeNumbers << " ";        for (int i = 0; i < NodeNumbers - 1; i++)        {            cout << fixed << setprecision(1) << Tmp->Exp << " " << fixed<<setprecision(1)<<Tmp->Coe << " ";            Tmp = Tmp->Next;        }        cout << fixed << setprecision(1) << Tmp->Exp << " " << fixed << setprecision(1) << Tmp->Coe<<endl;    }    friend void Product(Stack& S1,Stack& S2,Stack& Res);    void MergePolynomial()//合并同类项    {        List P = L->Next;        /*if (P->Next == NULL)            return;*/        while (P->Next!= NULL)        {            if (P->Exp == P->Next->Exp)            {                List Tmp = P->Next;                P->Coe = P->Coe + Tmp->Coe;                P->Next = Tmp->Next;                delete Tmp;                NodeNumbers--;            }            P = P->Next;            if (P == NULL)                break;        }    }    void ExpDescend()    {        if (IsEmpty())        {            cout << "Empty stack!" << endl;            return;        }        int* ExpArray = new int[NodeNumbers];        List LCopy = L->Next;        for (int i = 0; i < NodeNumbers; i++)        {            ExpArray[i] = LCopy->Exp;            LCopy = LCopy->Next;        }        qsort(ExpArray, NodeNumbers, sizeof(int), cmp);        List DescendList = new LinkedList;        DescendList->Coe = DescendList->Exp = 0;        DescendList->Next = NULL;        for (int i = 0; i < NodeNumbers; i++)        {            List P = L->Next;            while (P != NULL&&P->Exp != ExpArray[i])                P = P->Next;            if (P != NULL)            {                List TmpCell = new LinkedList;                TmpCell->Coe = P->Coe;                TmpCell->Exp = P->Exp;                TmpCell->Next = NULL;                TmpCell->Next = DescendList->Next;                DescendList->Next = TmpCell;                P->Exp = -1;            }        }        L = DescendList;    }    void DeleteNonTerm()    {        if (IsEmpty())        {            cout << "Empty stack!" << endl;            return;        }        List P = L;        while (P->Next != NULL)        {            if (abs(P->Next->Coe) < pow(10, -6))            {                List Temp = P->Next;                P->Next = Temp->Next;                delete Temp;                NodeNumbers--;            }            P = P->Next;        }    }};void Product(Stack& S1, Stack& S2,Stack& Res){    List Poly1 = S1.L;    List Poly2 = S2.L;    List ProdPoly = Res.L;    if (Poly1 == NULL)    {        cout << "Sorry, Poly1 Is NULL" << endl;        return;    }    if (Poly2 == NULL)    {        cout << "Sorry, Poly2 Is NULL" << endl;        return;    }    List P1 = Poly1->Next;//P1遍历Poly1    List P2 = Poly2->Next;//P2遍历Poly2    List P3 = ProdPoly;//P3在ProdPoly上移动    while (P1!=NULL)    {        while (P2 != NULL)        {            List P = (List)malloc(sizeof(LinkedList));            P->Coe = P1->Coe*P2->Coe;            P->Exp = P1->Exp + P2->Exp;            P->Next = NULL;            P3->Next = P;            P3 = P3->Next;            P2 = P2->Next;            Res.NodeNumbers++;        }        P1 = P1->Next;        P2 = Poly2->Next;    }    Res.ExpDescend();    Res.MergePolynomial();    Res.DeleteNonTerm();}int main(){    int K1, K2;    cin >> K1;    int* ExpArray1 = new int[K1];    double* CoeArray1 = new double[K1];    for (int i = 0; i < K1; i++)        cin >> ExpArray1[i] >> CoeArray1[i];    cin >> K2;    int* ExpArray2 = new int[K2];    double* CoeArray2 = new double[K2];    for (int i = 0; i < K2; i++)        cin >> ExpArray2[i] >> CoeArray2[i];    Stack S1, S2;    for (int i = 0; i < K1; i++)    {        S1.Push(ExpArray1[i], CoeArray1[i]);    }    for (int i = 0; i < K2; i++)    {        S2.Push(ExpArray2[i], CoeArray2[i]);    }    Stack Res;    Product(S1, S2, Res);    Res.Show();    delete[]ExpArray1;    delete[]CoeArray1;    delete[]ExpArray2;    delete[]CoeArray2;    return 0;}