PAT-A1009 Product of Polynomials 题目内容及题解
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 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
题目大意
题目给出A和B两个多项式,求出其积多项式并按照规定格式输出。
解题思路
- 分别读入两个多项式;
- 两个多项式各项依次相乘,并对指数相同的项的系数做合并;
- 按照题目要求从大到小输出多项式并返回0值。
代码
#include<stdio.h>
#include<math.h>
#define err 1E-7
#define MAXN 1005
struct POLY{
int K;
double cof[MAXN];
int exp[MAXN];
}a,b;
int main(){
int i,j,num=0,max;
double cof,res[2*MAXN];
int exp;
scanf("%d",&a.K);
for(i=0;i<a.K;i++){
scanf("%d%lf",&a.exp[i],&a.cof[i]);
}
scanf("%d",&b.K);
for(i=0;i<b.K;i++){
scanf("%d%lf",&b.exp[i],&b.cof[i]);
}
max=a.exp[0]+b.exp[0];
for(i=0;i<=max;i++){
res[i]=0.0;
}
for(i=0;i<a.K;i++){
for(j=0;j<b.K;j++){
exp=a.exp[i]+b.exp[j];
cof=a.cof[i]*b.cof[j];
if(fabs(res[exp])<err){
res[exp]=cof;
num++;
}else{
res[exp]+=cof;
if(fabs(res[exp])<err){
num--;
}
}
}
}
printf("%d ",num);
for(i=max;i>=0;i--){
if(fabs(res[i])>err){
printf("%d %.1lf",i,res[i]);
num--;
if(num>0){
printf(" ");
}else{
printf("\n");
}
}
}
return 0;
}
运行结果