1002 A+B for Polynomials (25)（25 point(s)）

problem

1002 A+B for Polynomials (25)（25 point(s)）

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

anwser

#include<iostream>

#include<stdio.h>

#include<iomanip>

#include<cstring>

int main()

{

	freopen("test.txt", "r",stdin);

	float n[1001], n1[1001], n2[1001];

	memset(n, 0, 1001*sizeof(float));

	memset(n1, 0, 1001*sizeof(float));

	memset(n2, 0, 1001*sizeof(float));

	int a, b;

	std::cin>>a;

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

		int temp1;

		float temp2;

		std::cin>>temp1>>temp2;

		n1[temp1] = temp2;

	}

	std::cin>>b;

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

		int temp1;

		float temp2;

		std::cin>>temp1>>temp2;

		n2[temp1] = temp2;

	}

	int c = 0;

	for(int i = 0; i < 1001; i++){

		n[i] = n1[i] + n2[i];

//		std::cout<<n[i]<<i<<std::endl;

		if(n[i] != 0) c++;

	}

	std::cout<<c;

	for(int i = 1000; i >= 0; i--){

		if(n[i] != 0) {

			std::cout<<" "<<i<<" ";

			std::cout<<std::fixed<<std::setprecision(1)<<n[i];

//			printf("%.1f", n[i]);

		}

	}

	return 0;

}

/*

2 1 2.4 0 3.2

2 2 1.5 1 0.5

*/

experience

注意头文件名
注意边界条件以及输出格式

单词复习：

polynomials 多项式
exponents 范例，指数
coefficients 系数
respectively 分别的
accurate 精确的
decimal 小数

巴特西

1002 A+B for Polynomials (25)（25 point(s)）

problem

anwser

experience

最新文章

热门文章