http://pat.zju.edu.cn/contests/pat-practise/1009

 1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

一开始犯错误了，想乘结果不为0的后面有可能为0，负+正=0。所以还是在最后判断项是否为0比较简单。

[cpp] view plaincopy

1. #include<iostream>  
2. #include<cstdio>  
3. #include<memory.h>  
4. #include<algorithm>  
5. #include<cstring>  
6. #include<queue>  
7. #include<cmath>  
8. #include<cstdlib>  
9. #include<string>  
10. using namespace std;  
11. double ak[1001];  
12. double bk[1001];  
13. double ck[2002];  
14. int main(){  
15.   
16.     //freopen("in.txt", "r", stdin);  
17.   int n;  
18.   int a;  
19.   double b;  
20.   while(cin>>n){  
21.     memset(ak, 0, sizeof(ak));  
22.     memset(bk, 0, sizeof(bk));  
23.     memset(ck, 0, sizeof(ck));  
24.     for(int i=0;i<n;++i){  
25.       scanf("%d %lf", &a, &b);  
26.       ak[a] = b;  
27.     }  
28.     cin>>n;  
29.     for(int i=0;i<n;++i){  
30.       scanf("%d %lf", &a, &b);  
31.       bk[a] = b;  
32.     }  
33.     int cnt = 0;  
34.     for(int i=0;i<1001;++i)  
35.       for(int j=0;j<1001;++j){  
36.         if(ak[i]!=0 && bk[j]!=0){  
37.           ck[i+j] += ak[i]*bk[j];//+=  
38.         }  
39.       }  
40.     /*最后判断比较简单*/  
41.     for(int i=0;i<2002;++i)  
42.       if(ck[i]!=0)  
43.         cnt++;  
44.     printf("%d", cnt);  
45.     for(int i=2001;i>=0;i--){  
46.       if(ck[i]!=0)  
47.         printf(" %d %.1lf", i, ck[i]);  
48.     }  
49.     printf("\n");    
50.       
51.   }  
52.   
53.     //fclose(stdin);  
54.     return 0;  
55. }