1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

65536 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

#include<stdio.h>#include<malloc.h>#include<iostream>#include<map>#include<vector>#include<algorithm>using namespace std;typedef struct LinkNode *link;struct LinkNode{int expo;double index;}; map<int ,double> m;LinkNode number1[10];LinkNode number2[10];vector<LinkNode> v;bool compare1(LinkNode a,LinkNode b){return a.expo>b.expo;}int main(){int N,M;cin>>N;int i,j;int exp;double index;map<int ,double>::iterator it;for(i=0;i<N;i++){cin>>number1[i].expo>>number1[i].index;}cin>>M;for(i=0;i<M;i++){cin>>number2[i].expo>>number2[i].index;}for(i=0;i<N;i++){for(j=0;j<M;j++){it=m.find(number1[i].expo+number2[j].expo);if(it!=m.end())it->second+=number1[i].index*number2[j].index;elsem.insert(pair<int,double>(number1[i].expo+number2[j].expo,number1[i].index*number2[j].index));}}//sort(m.begin(),m.end(),compare1);for(it=m.begin();it!=m.end();it++){if(it->second!=0){LinkNode l;l.expo=it->first;l.index=it->second;v.push_back(l);}}sort(v.begin(),v.end(),compare1);cout<<v.size();for(i=0;i<v.size();i++){printf(" %d %.1f",v[i].expo,v[i].index);}}

感想：这道题最好是用map，然后把非零项添加到线性容器vector中，然后再自定义排序就可以出来