1009. 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

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 
 5 double str1[1001],str2[1001],ans[2002];
 6 int main()
 7 {
 8     int n,m,zi,xi,cnt=0;
 9     int i,j;
10     scanf("%d",&n);
11     for( i=0; i<n; i++)
12     {
13         scanf("%d",&zi);
14         scanf("%lf",&str1[zi]);
15     }
16     scanf("%d",&m);
17     for( i=0; i<m; i++)
18     {
19         scanf("%d",&zi);
20         scanf("%lf",&str2[zi]);
21     }
22     for( i=0; i<1001; i++)
23     {
24         if( str1[i])
25         {
26             for( j=0; j<1001; j++)
27                 if( str2[j])
28                     ans[i+j] += str1[i]*str2[j];
29         }
30     }
31     for( i=0; i<2001; i++)
32         if( ans[i])
33             cnt++;
34     printf("%d",cnt);
35     for( i=2001; i>=0; i--)
36     {
37         if( ans[i])
38             printf(" %d %.1f",i,ans[i]);
39     }
40     return 0;
41 }

 

posted @ 2018-03-11 10:35  yuxiaoba  阅读(121)  评论(0)  编辑  收藏  举报