pat甲级1002 A+B for Polynomials (多项式加法)

1002 A+B for Polynomials (25 分)

This time, you are supposed to find A+B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

K N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

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

解题思路

这题是求多项式的加法,第一个数是次数,第二个数是系数,因为次数不超过1000,所以开个double的数组a[1005]和b[1005]初始化为0,a[]存一个多项式,b[]存第二个,a[i]的值表示次数为i的项的系数。相同次数加在一起就行了,输出按照次数从大到小,输出系数不为0的项。

代码如下

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
double a[1005], b[1005];
double c[1005];
int main()
{
	int k1;
	while(cin >> k1){
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		for(int i = 0; i < k1; i ++){
			int x;
			double y;
			cin >> x >> y;
			a[x] += y;
		}
		int k2;
		cin >> k2;
		for(int i = 0; i < k2; i ++){
			int x;
			double y;
			cin >> x >> y;
			b[x] += y;
		}
		for(int i = 0; i <= 1000; i ++){
			c[i] = a[i] + b[i];
		}		
		int cnt = 0;
		for(int i = 0; i <= 1000; i ++){
			if(c[i]){
				cnt ++;
			}
		}
		cout << cnt;
		for(int i = 1000; i >= 0; i --){
			if(c[i])
				printf(" %d %.1lf", i, c[i]);
		}
		cout << endl;
	}
	return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值