PATA 1002. A+B for Polynomials (25) 解题报告

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

题目解释：

在两行分别给出两个多项式，在输入的每一行中，先给出一个数字K，代表该行非零项的个数，然后接下来一次给出每一项的指数和系数，本例的输入样例是2.4X+3.2，第二行是1.5X^2+0.5X，相加的结果为1.4

AC代码

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int k;
	int count=0;
	double data[1005]={0};
	int exp;
	double coe;
	cin>>k;
	for(int i=0;i<k;i++)//巧妙的保存方法 
	{
		cin>>exp>>coe;
		data[exp]+=coe;
	}
	cin>>k;
	for(int i=0;i<k;i++)//把具有相同指数的项系数相加 
	{
		cin>>exp>>coe;
		data[exp]+=coe;
	}
	for(int i=1004;i>=0;i--)//根据题目描述，指数最高是1000，于是我们设置一个循环检索非零项的个数 
	{
		if(data[i]!=0)
		count++;
	}
	cout<<count;
	for(int i=1004;i>=0;i--)//按照多项式的约定，从高次项低次输出 
	{
		if(data[i]!=0)
		printf(" %d %.1f",i,data[i]);
	}
	return 0;
} 

总结：本题的关键是用了按指数保存系数的方法，同时还要注意输出的格式，按照多项式的约定从高次向低次输出，注意控制空格