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

View Code

  1 #include<stdio.h>

  2 struct node

  3 {

  4     int exponents;

  5     float coefficients;

  6 };

  7 struct Polynomials{

  8     int number;

  9     node items[15];

 10     Polynomials(){

 11         number = 0;

 12     }

 13 };

 14 struct product {

 15     int number;

 16     node items[105];

 17     product(){

 18         number = 0;

 19     }

 20 };

 21 void insertItem(product &a, const node &b)

 22 {

 23     int i,j;

 24     // a 为空

 25     if (a.number == 0)

 26     {

 27         a.items[0] = b ;

 28         a.number ++;

 29         return ;

 30     }

 31     // 插在末尾后面

 32     if (a.items[a.number-1].exponents> b.exponents)

 33     {

 34         a.items[a.number] = b;

 35         a.number++;

 36         return ;

 37     }

 38 

 39     //寻找插的位置

 40     for (i = a.number-1; i >= 0; i--)

 41     {

 42         if (a.items[i].exponents<b.exponents)

 43              continue;

 44         else

 45             break;    

 46     }

 47 

 48     //插在最前面

 49     if(i == -1){

 50         j = a.number-1;

 51 

 52         while(j>=0){

 53             a.items[j+1] =a.items[j];

 54             j--;

 55         }

 56         a.items[0] = b;

 57         a.number++;

 58         return ;

 59     }

 60 

 61     //插在中间

 62     if(a.items[i].exponents == b.exponents){

 63         

 64         a.items[i].coefficients+= b.coefficients ;

 65         //两项和为零

 66         if (a.items[i].coefficients == 0)

 67         {

 68             j = i;

 69             while (j < a.number -1)

 70             {

 71                 a.items[j] = a.items[j+1];

 72             }

 73             a.number --;

 74             return ;

 75         }

 76     }else{

 77         

 78         j = a.number-1;

 79 

 80         while (j>i)

 81         {

 82             a.items[j+1] =a.items[j];

 83             j--;

 84         }

 85 

 86         a.items[i+1] = b;

 87         a.number++;

 88         return ;

 89 

 90     }

 91 

 92 }

 93 void insert(product &a,const Polynomials & b){

 94     

 95     int i;

 96     // 初始化

 97     if(a.number == 0){

 98         a.number = b.number;

 99         for(i = 0; i < b.number; i++)

100             a.items[i] = b.items[i];

101         return ;

102     }

103     for (i = 0; i <b.number ; i++)

104           insertItem(a, b.items[i]);

105 

106 }

107 Polynomials p1,p2,p3;

108 product  result;

109 int main()

110 {

111     int i,j,k;

112     int exponents;

113     float coefficients;

114     while (scanf("%d", &k) !=EOF){

115     

116     result.number = 0;

117     p1.number = k;

118     for (i =0; i<k;i++)

119     {

120         scanf("%d %f",&p1.items[i].exponents, &p1.items[i].coefficients);    

121     }

122 

123 

124     scanf("%d", &k);

125     p2.number = k;

126     for (i =0; i<k;i++)

127     {

128         scanf("%d %f",&p2.items[i].exponents, &p2.items[i].coefficients);    

129     }

130 

131     p3.number =p1.number;

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

133 

134         for (j =0; j<p1.number ;j++)

135         {

136             p3.items[j].exponents = p1.items[j].exponents + p2.items[i].exponents;

137             p3.items[j].coefficients = p1.items[j].coefficients * p2.items[i].coefficients ;

138         }

139         insert(result, p3);

140         

141     }

142 

143     printf("%d", result.number) ;

144     for (i = 0; i< result.number ;i++)

145     {

146         printf(" %d %.1f", result.items[i].exponents, result.items[i].coefficients);

147     }

148     

149     }

150     return 0;

151 }