pat 1009（需回顾）

Product of Polynomials (25)

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

---

这道题难倒不难，就是细节需要注意的多点，主要有两方面： 一，每一项插入以计算出的积时，有三个地方需要考虑，头 尾 中间;二 结果为零的需要删除节点

#include<stdio.h>
struct node
{
    int exponents;
    float coefficients;
};
struct Polynomials{
    int number;
    node items[15];
    Polynomials(){
        number = 0;
    }
};
struct product {
    int number;
    node items[105];
    product(){
        number = 0;
    }
};
void insertItem(product &a, const node &b)
{
    int i,j;
    // a 为空
    if (a.number == 0)
    {
        a.items[0] = b ;
        a.number ++;
        return ;
    }
    // 插在末尾后面
    if (a.items[a.number-1].exponents> b.exponents)
    {
        a.items[a.number] = b;
        a.number++;
        return ;
    }

    //寻找插的位置
    for (i = a.number-1; i >= 0; i--)
    {
        if (a.items[i].exponents<b.exponents)
             continue;
        else
            break;    
    }

    //插在最前面
    if(i == -1){
        j = a.number-1;

        while(j>=0){
            a.items[j+1] =a.items[j];
            j--;
        }
        a.items[0] = b;
        a.number++;
        return ;
    }

    //插在中间
    if(a.items[i].exponents == b.exponents){
        
        a.items[i].coefficients+= b.coefficients ;
        //两项和为零
        if (a.items[i].coefficients == 0)
        {
            j = i;
            while (j < a.number -1)
            {
                a.items[j] = a.items[j+1];
            }
            a.number --;
            return ;
        }
    }else{
        
        j = a.number-1;

        while (j>i)
        {
            a.items[j+1] =a.items[j];
            j--;
        }

        a.items[i+1] = b;
        a.number++;
        return ;

    }

}
void insert(product &a,const Polynomials & b){
    
    int i;
    // 初始化
    if(a.number == 0){
        a.number = b.number;
        for(i = 0; i < b.number; i++)
            a.items[i] = b.items[i];
        return ;
    }
    for (i = 0; i <b.number ; i++)
          insertItem(a, b.items[i]);

}
Polynomials p1,p2,p3;
product  result;
int main()
{
    int i,j,k;
    int exponents;
    float coefficients;
    while (scanf("%d", &k) !=EOF){
    
    result.number = 0;
    p1.number = k;
    for (i =0; i<k;i++)
    {
        scanf("%d %f",&p1.items[i].exponents, &p1.items[i].coefficients);    
    }


    scanf("%d", &k);
    p2.number = k;
    for (i =0; i<k;i++)
    {
        scanf("%d %f",&p2.items[i].exponents, &p2.items[i].coefficients);    
    }

    p3.number =p1.number;
    for(i =0;i <p2.number ;i++){

        for (j =0; j<p1.number ;j++)
        {
            p3.items[j].exponents = p1.items[j].exponents + p2.items[i].exponents;
            p3.items[j].coefficients = p1.items[j].coefficients * p2.items[i].coefficients ;
        }
        insert(result, p3);
        
    }

    printf("%d", result.number) ;
    for (i = 0; i< result.number ;i++)
    {
        printf(" %d %.1f", result.items[i].exponents, result.items[i].coefficients);
    }
    
    }
    return 0;
}