1009 Product of Polynomials (25分)
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 N?1?? a?N?1???? N?2?? a?N?2???? ... N?K?? a?N?K????
where K is the number of nonzero terms in the polynomial, N?i?? and a?N?i???? (i=1,2,?,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N?K??<?<N?2??<N?1??≤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
题意解析:通过结构体,将两个多项式的信息储存起来。将两个多项式相乘,结果储存到DOUBLE数组当中(便于访问)。当两个多项式相乘完毕后,最后再遍历非零的数(我在最初的计算过程中并未考虑到多项式相乘正负相反想加刚好等0的情况,和结果为符的情况),最后遍历就不用统计结果为零的情况,切记统计结果为负的情况,第一个检测点就有负数的情况。
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
struct Number{
int n;
double a;
}A[11],B[11];
double C[2002]={0};
int main(){
int K1,n,K2,c1,minnum,N;
double a,c2;
minnum=2001,N=0;
scanf("%d ",&K1);
for(int i=0;i<K1;i++){
scanf("%d %lf",&n,&a);
A[i].n=n;
A[i].a=a;
}
scanf("%d ",&K2);
for(int i=0;i<K2;i++){
scanf("%d %lf",&n,&a);
B[i].n=n;
B[i].a=a;
}
for(int i=0;i<K1;i++){
for(int j=0;j<K2;j++){
c1=A[i].n+B[j].n;
c2=A[i].a*B[j].a;
C[c1]=C[c1]+c2;
}
}
for(int i=0;i<2002;i++){
if(C[i]!=0){
N++;
}
}
printf("%d",N);
for(int i=2002;i>=0;i--){
if(C[i]!=0){
printf(" %d %.1f",i,C[i]);
}
}
}