【PAT】1009. Product of Polynomials (25)

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1009

分析：简单题。相乘时指数相加，系数相乘即可，输出时按指数从高到低的顺序。注意点：多项式相乘后指数最高可达2000。

题目描述：

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 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

参考代码：

#include<iostream>
#include<iomanip>
#include<string.h>
using namespace std;#define max 1000
double input1[max + 1];
double input2[max + 1];
double result[2*max + 1];int main()
{memset(input1,0,sizeof(input1));memset(input2,0,sizeof(input2));memset(result,0,sizeof(result));int k;int i,j;int e;//int temp=0; //记录最高指数double c;int count = 0; //最终输出的多项式的项数。cin>>k;for(i=0; i<k; i++){cin>>e>>c;input1[e] += c;}cin>>k;for(i=0; i<k; i++){cin>>e>>c;input2[e] += c;}for(i=0; i<=1000; i++){for(j=0; j<=1000; j++){result[i+j] += input1[i]*input2[j];}}for(i=0; i<=2000; i++) if(result[i] != 0) count++;cout<<count;for(i=2000; i >= 0; i--){if(result[i] != 0.0) {cout<<" "<<i;cout<<fixed<<setprecision(1);cout<<" "<<result[i];}}cout<<endl;return 0;
}

转载于:https://www.cnblogs.com/pangblog/p/3279789.html

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