http://pat.zju.edu.cn/contests/pat-practise/1009-优快云博客

本文链接：https://blog.youkuaiyun.com/cstur4/article/details/8069307

本文介绍了一种用于计算两个多项式乘积的方法，并详细解释了如何通过输入多项式的系数和指数，输出它们的乘积。算法适用于多项式数量不超过10且每个项的指数不超过1000的情况。

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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 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 Input2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output3 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 
      
 #include<iostream>  
 #include<cstdio>  
 #include<memory.h>  
 #include<algorithm>  
 #include<cstring>  
 #include<queue>  
 #include<cmath>  
 #include<cstdlib>  
 #include<string>  
 using namespace std;  
 double ak[1001];  
 double bk[1001];  
 double ck[2002];  
 int main(){  
   
     //freopen("in.txt", "r", stdin);  
   int n;  
   int a;  
   double b;  
   while(cin>>n){  
     memset(ak, 0, sizeof(ak));  
     memset(bk, 0, sizeof(bk));  
     memset(ck, 0, sizeof(ck));  
     for(int i=0;i<n;++i){  
       scanf("%d %lf", &a, &b);  
       ak[a] = b;  
     }  
     cin>>n;  
     for(int i=0;i<n;++i){  
       scanf("%d %lf", &a, &b);  
       bk[a] = b;  
     }  
     int cnt = 0;  
     for(int i=0;i<1001;++i)  
       for(int j=0;j<1001;++j){  
         if(ak[i]!=0 && bk[j]!=0){  
           ck[i+j] += ak[i]*bk[j];//+=  
         }  
       }  
     /*最后判断比较简单*/  
     for(int i=0;i<2002;++i)  
       if(ck[i]!=0)  
         cnt++;  
     printf("%d", cnt);  
     for(int i=2001;i>=0;i--){  
       if(ck[i]!=0)  
         printf(" %d %.1lf", i, ck[i]);  
     }  
     printf("\n");    
       
   }  
   
     //fclose(stdin);  
     return 0;  
 }  