PAT A1002 A+B for Polynomials(25)

1002 A+B for Polynomials (25)（25 分）

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 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (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.//保留一位小数,最后一个不能有空格

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

思路:

        输入一个数字,表示多项式的个数,然后输入指数(e)和系数(c),将他们保存在一个数组中,用下标来表示指数(e)

这个就表示的是: 2.4x+3.2

采用这种方式将输入的值存到数组中,题目要求从按照指数从大到小输出,那就将数组倒序输出,判断a[e]为0,则不输出,仅仅输出那些非0的printf("%d%.1lf",i,a[i]);

AC代码:

#include <stdio.h>
int main(){
	int N,M,count=0,e,t=10;
	double c;
	double a[1111]={0},b[1111]={0},res[1111]={0};
	scanf("%d",&N);
	for(int i=0;i<N;i++){
		scanf("%d %lf",&e,&c);
		a[e]=c;//下标即指数
	}
	
	scanf("%d",&M);
	for(int i=0;i<M;i++){
		scanf("%d %lf",&e,&c);
		b[e]=c;
	}
	for(int i=0;i<1111;i++){//将两个多项式,按照下标对应相加,(即按照指数对应相加)
		res[i]=a[i]+b[i];    
		if(res[i]!=0) count++;//统计多项式系数非0个数
	}
	printf("%d",count);//输出多项式系数非0个数
	for(int i=1110;i>=0;i--){
		if(res[i]!=0){//只输出非0
			printf(" %d %.1f",i,res[i]);
		}
	}
	return 0;
}

summary:

            其实也可以只用一个数组存放:

for(int i=0;i<N;i++){

        scanf("%d %lf",&e,&c);

      a[e]+=c;//把下标(指数)相同的存在一个数组进行累加就可以了

}