PAT-A1002 A+B for 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 sum 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 to 1 decimal place.
Sample Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output:
3 2 1.5 1 2.9 0 3.2
题目大意
题目给出A和B两个多项式,求出其和多项式并按照规定格式输出。
解题思路
- 读入第一个多项式,并时刻保持记录当前最大的指数;
- 依次读入第二个多项式,并根据其内容做出修正(需注意double判别相等不能直接使用‘==’);
- 按照题目要求从大到小输出多项式并返回零值。
代码
#include<stdio.h>
#include<math.h>
#define maxn 1010
#define err 1E-7
int main(){
double a[maxn]={0.0};
int num=0,max=0;//max记录当前最大的指数值
int k,i,exp;
double coe;
scanf("%d",&k);
num=k;
for(i=0;i<k;i++){
scanf("%d%lf",&exp,&coe);
a[exp]=coe;
max=(exp>max?exp:max);
}
scanf("%d",&k);
for(i=0;i<k;i++){
scanf("%d%lf",&exp,&coe);
if(fabs(a[exp])<err){
max=(exp>max?exp:max);
num++;
}
a[exp]=a[exp]+coe;
if(fabs(a[exp])<err){
num--;
}
}
printf("%d",num);
while(max>=0){
if(fabs(a[max])>=err){
printf(" %d %.1f",max,a[max]);
num--;
}
if(num<=0){
break;
}
max--;
}
printf("\n");
return 0;
}
运行结果