1002. A+B for Polynomials (25)_1002 a+b for polynomials (25分 scanf("%d",k); for(i-优快云博客

本文介绍了一个计算两个多项式相加的程序实现。输入包括两个多项式的系数和指数，输出为两个多项式相加的结果。程序使用了数组来存储每个指数对应的系数，并最终输出结果。

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1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

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

#include<stdio.h>
double e[1010];
int main(){
	int k,i,exp;
	double coe;
	scanf("%d",&k);
	for(i=0;i<k;i++){
		scanf("%d %lf",&exp,&coe);
		e[exp]=e[exp]+coe;
	}
	scanf("%d",&k);
	for(i=0;i<k;i++){
		scanf("%d %lf",&exp,&coe);
		e[exp]=e[exp]+coe;
	}
	int cou=0;
	for(i=0;i<=1000;i++){
		if(e[i]!=0){
			cou++;
		}
	}
	printf("%d",cou);
	for(i=1000;i>=0;i--){
		if(e[i]!=0){
			printf(" %d %.1lf",i,e[i]);
		}
	}
}