1002. A+B for Polynomials (25)-PAT Grade A Zhenti
Details：This time, you are supposed to find A+B where A and B are two polynomials.
Input
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
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
The main idea of the title: Calculate the sum of the polynomials A+B (given in descending order of exponents), and output the exponents and coefficients whose coefficients are not 0 in descending order
Analysis of ideas: set up a two-dimensional array, put one coefficient and one exponent, and then accept a and b. Add the corresponding exponent to c (the same exponent coefficients are added directly), and then output the exponent sum with non-zero coefficients. coefficient
Codes
#include<cstdio>
int main(){
int k,m,d,count=0;
double a[10][2],b[10][2],c[20][2]={0.0};
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d %lf",&d,&a[i][1]);
a[i][0]=d;
}
scanf("%d",&m);
for(int i=0;i<m;i++){
scanf("%d %lf",&d,&b[i][1]);
b[i][0]=d;
}
int x=0,y=0;
while(x<k&&y<m){
if(a[x][0]>b[y][0]){
c[count][0]=a[x][0];
c[count][1]=a[x][1];
x++;
if(c[count][1]!=0)
{
count++;
}
}
else if(a[x][0]==b[y][0]){
c[count][0]=a[x][0];
c[count][1]=a[x][1]+b[y][1];
x++;
y++;
if(c[count][1]!=0)
{
count++;
}
}
else{
c[count][0]=b[y][0];
c[count][1]=b[y][1];
y++;
if(c[count][1]!=0)
{
count++;
}
}
}
while(x<k){
c[count][0]=a[x][0];
c[count][1]=a[x][1];
x++;
if(c[count][1]!=0)
{
count++;
}
}
while(y<m){
c[count][0]=b[y][0];
c[count][1]=b[y][1];
y++;
if(c[count][1]!=0)
{
count++;
}
}
printf("%d",count);
for(int i=0;i<20;i++){
if(c[i][1]!=0){
printf(" %d %.1f",(int)c[i][0],c[i][1]);
}
}
return 0;
}