Problem Description

要求采用链表形式，求两个一元多项式的乘积：h3 = h1*h2。函数原型为：void multiplication( NODE * h1, NODE * h2, NODE * h3 )。

输入：

输入数据为两行，分别表示两个一元多项式。每个一元多项式以指数递增的顺序输入多项式各项的系数（整数）、指数（整数）。

例如：1+2x+x2表示为：<1,0>,<2,1>,<1,2>,

输出：

以指数递增的顺序输出乘积： <系数,指数>,<系数,指数>,<系数,指数>,

零多项式的输出格式为：<0,0>,

测试输入

<1,0>,<2,1>,<1,2>,
<1,0>,<1,1>,

测试输出

<1,0>,<3,1>,<3,2>,<1,3>,

预置代码

/* PRESET CODE BEGIN - NEVER TOUCH CODE BELOW */  
 
#include <stdio.h>  
#include <stdlib.h>  
  
typedef struct node  
{   int    coef, exp;  
    struct node  *next;  
} NODE;  
  
void multiplication( NODE *, NODE * , NODE * );  
void input( NODE * );  
void output( NODE * );  
  
void input( NODE * head )  
{   int flag, sign, sum, x;  
    char c;  
  
    NODE * p = head;  
  
    while ( (c=getchar()) !='\n' )  
    {  
        if ( c == '<' )  
        {    sum = 0;  
             sign = 1;  
             flag = 1;  
        }  
        else if ( c =='-' )  
             sign = -1;  
        else if( c >='0'&& c <='9' )  
        {    sum = sum*10 + c - '0';  
        }  
        else if ( c == ',' )  
        {    if ( flag == 1 )  
             {    x = sign * sum;  
                  sum = 0;  
                  flag = 2;  
          sign = 1;  
             }  
        }  
        else if ( c == '>' )  
        {    p->next = ( NODE * ) malloc( sizeof(NODE) );  
             p->next->coef = x;  
             p->next->exp  = sign * sum;  
             p = p->next;  
             p->next = NULL;  
             flag = 0;  
        }  
    }  
}  
  
void output( NODE * head )  
{  
    while ( head->next != NULL )  
    {   head = head->next;  
        printf("<%d,%d>,", head->coef, head->exp );  
    }  
    printf("\n");  
}  
  
int main()  
{   NODE * head1, * head2, * head3;  
  
    head1 = ( NODE * ) malloc( sizeof(NODE) );  
    input( head1 );  
  
    head2 = ( NODE * ) malloc( sizeof(NODE) );  
    input( head2 );  
  
    head3 = ( NODE * ) malloc( sizeof(NODE) );  
    head3->next = NULL;  
    multiplication( head1, head2, head3 );  
  
    output( head3 );  
  
    return 0;  
}  
  
/* PRESET CODE END - NEVER TOUCH CODE ABOVE */  

AcCode

//
//  main.cpp
//  一元多项式相乘
//
//  Created by jetviper on 2017/3/26.
//  Copyright © 2017年 jetviper. All rights reserved.
//

/* PRESET CODE BEGIN - NEVER TOUCH CODE BELOW */

#include <stdio.h>
#include <stdlib.h>

typedef struct node
{   int    coef, exp;
    struct node  *next;
} NODE;

void multiplication( NODE *, NODE * , NODE * );
void input( NODE * );
void output( NODE * );

void input( NODE * head )
{   int flag, sign, sum, x;
    char c;
    
    NODE * p = head;
    
    while ( (c=getchar()) !='\n' )
    {
        if ( c == '<' )
        {    sum = 0;
            sign = 1;
            flag = 1;
        }
        else if ( c =='-' )
            sign = -1;
        else if( c >='0'&& c <='9' )
        {    sum = sum*10 + c - '0';
        }
        else if ( c == ',' )
        {    if ( flag == 1 )
        {    x = sign * sum;
            sum = 0;
            flag = 2;
            sign = 1;
        }
        }
        else if ( c == '>' )
        {    p->next = ( NODE * ) malloc( sizeof(NODE) );
            p->next->coef = x;
            p->next->exp  = sign * sum;
            p = p->next;
            p->next = NULL;
            flag = 0;
        }
    }
}

void output( NODE * head )
{
    while ( head->next != NULL )
    {   head = head->next;
        printf("<%d,%d>,", head->coef, head->exp );
    }
    printf("\n");
}

int main()
{   NODE * head1, * head2, * head3;
    
    head1 = ( NODE * ) malloc( sizeof(NODE) );
    input( head1 );
    
    head2 = ( NODE * ) malloc( sizeof(NODE) );
    input( head2 );
    
    head3 = ( NODE * ) malloc( sizeof(NODE) );
    head3->next = NULL;
    multiplication( head1, head2, head3 );
    
    output( head3 );
    
    return 0;
}

/* PRESET CODE END - NEVER TOUCH CODE ABOVE */

void multiplication(NODE * h1, NODE * h2, NODE * h3)
{
    NODE *p1 = h2, *p2 = h1, *p3 = h3;
    NODE *temp,*t1;
    int cl = 0;
    
    
    
    while (p1->next != NULL)
    {
        p2 = h1;
        p1 = p1->next;
        if (p1->coef == 0&&cl==1)continue;
        p3 = h3;
        while (p2->next != NULL)
        {
            p2 = p2->next;
            int zs = p1->exp +p2->exp ;
            
            int xs = p1->coef * p2->coef;
            
            if (xs == 0 && cl == 1)continue;
            
            
            NODE *tp;
            while (p3!=NULL) {
                
                tp = p3;
                p3 = p3->next;
                if (p3 == NULL) {
                    temp = (NODE*)malloc(sizeof(NODE));
                    if (xs == 0&&cl==0) {
                        temp->exp = 0;
                        cl = 1;
                    }
                    else temp->exp = zs;
                    temp->coef = xs;
                    temp->next = p3;
                    tp->next = temp;
                    p3 = tp;
                    break;
                }
                if (p3->exp < zs)continue;
                if (p3->exp == zs) {
                    p3->coef += xs;
                    if (p3->coef == 0 && zs != 0) {
                        tp->next = p3->next;
                    }
                    p3 = tp;
                    break;
                }
                if (p3->exp > zs) {
                    temp = (NODE*)malloc(sizeof(NODE));
                    if (xs == 0&&cl==0) {
                        temp->exp = 0;
                        cl = 1;
                    }
                    else temp->exp = zs;
                    temp->coef = p1->coef * p2->coef;
                    temp->next = p3;
                    tp->next = temp;
                    p3 = tp;
                    break;
                }
            }
        }
        
    }
}