PAT甲级 1009.Product of Polynomials(25) 题目翻译与答案

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题目来源自PAT网站 https://www.patest.cn/

题目描述：

1009. Product of Polynomials (25)

This time, you aresupposed to find A*B where A and B are two polynomials.

InputSpecification:

Each input filecontains one test case. Each case occupies 2 lines, and each line contains theinformation of a polynomial: K N1 a_N1 N2 a_N2 ...NK a_NK, where K is the number of nonzero terms in the polynomial, Niand a_Ni (i=1, 2, ..., K) are the exponents and coefficients,respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2< N1 <=1000.

OutputSpecification:

For each test caseyou should output the product of A and B in one line, with the same format asthe input. Notice that there must be NO extra space at the end of each line.Please be accurate up to 1 decimal place.

SampleInput

2 1 2.4 0 3.2

2 2 1.5 1 0.5

SampleOutput

3 3 3.6 2 6.0 11.6

题目翻译：

1009.多项式的乘积

这一次，你需要计算出A*B，而A与B是两个多项式。

输入说明：

每个输入文件包含一个测试实例。每个实例有两行，每行是一个多项式的信息K N1 a_N1 N2 a_N2 ... NKa_NK，K是多项式中非零项的个数，Ni和 a_Ni (i=1, 2, ..., K) 分别是指数和系数，且1 <= K <= 10, 0 <= NK < ... < N2 < N1<=1000。

输出说明：

对于每个测试实例，你应该在一行上输出A与B的乘积，格式同输入时一样。注意在每行的末尾不能有多余的空格。精确度到一位小数。

输入样例：

2 1 2.4 0 3.2

2 2 1.5 1 0.5

输出样例：

3 3 3.6 2 6.0 1 1.6

答案代码：

#include<cstdio>struct poly{int k;int exp[10];double coe[10];};struct polyPro{int k;int exp[100];double coe[100];};void inputPoly(poly &p){int i;scanf("%d",&(p.k));for(i=0;i<p.k;++i){scanf("%d",&(p.exp[i]));scanf("%lf",&(p.coe[i])); }}void outputPro(polyPro &p){int i;printf("%d",p.k);for(i=0;i<p.k;++i){if(p.coe[i]>-0.05 && p.coe[i]<0.05)continue;printf(" %d %.1lf",p.exp[i],p.coe[i]);}}void output(poly &p){//test functionint i;printf("%d ",p.k);for(i=0;i<p.k-1;++i)printf("%d %.1lf ",p.exp[i],p.coe[i]);printf("%d %.1lf\n",p.exp[i],p.coe[i]);}void exchange(polyPro &p,int i,int j){int ti;double td;ti=p.exp[i];p.exp[i]=p.exp[j];p.exp[j]=ti;td=p.coe[i];p.coe[i]=p.coe[j];p.coe[j]=td;}polyPro calPoly(poly &a,poly &b){polyPro p;int i1,i2,i3,i=0;int expT;double coeT;for(i1=0;i1<a.k;++i1)for(i2=0;i2<b.k;++i2){expT=a.exp[i1]+b.exp[i2];coeT=a.coe[i1]*b.coe[i2];for(i3=0;i3<i;++i3){if(expT==p.exp[i3]){p.coe[i3]+=coeT;coeT=0;}}if(coeT!=0){p.exp[i]=expT;p.coe[i]=coeT;i++;}}p.k=i;for(i1=0,i2=0;i1<p.k;i1++){if(p.coe[i1]>-0.05 && p.coe[i1]<0.05)i--;else{p.coe[i2]=p.coe[i1];p.exp[i2]=p.exp[i1];i2++;}}p.k=i2;int iMax;for(i1=0;i1<p.k-1;++i1){iMax=i1; for(i2=i1+1;i2<p.k;++i2){if(p.exp[iMax]<p.exp[i2])iMax=i2;}if(iMax!=i1)exchange(p,iMax,i1);}return p;}int main(){poly p1,p2;inputPoly(p1);inputPoly(p2);//output(p1); output(p2); polyPro p=calPoly(p1,p2);//printf("@@@@@@@@@@@@\n");outputPro(p); return 0;}

说明与心得：

我留下了代码的debug痕迹。

题目方法选择的不太好，尤其是输入有序这个条件我完全没有利用。反而用了选择排序。

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