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 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.
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 }