NBUT The Sum of F(x) and G(x)

最新推荐文章于 2023-03-06 19:36:40 发布

Joyyiwei

最新推荐文章于 2023-03-06 19:36:40 发布

阅读量862

点赞数

CC 4.0 BY-SA版权

分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/u011345136/article/details/24991649

搜索专栏收录该内容

138 篇文章

订阅专栏

本文介绍了一个游戏场景中的数学问题，玩家需要解决多项式的加法问题才能进入古墓。通过给出具体的例子，文章详细展示了如何将两个多项式相加并简化结果的过程。提供了完整的代码实现，帮助读者理解算法的具体步骤。

问题描述
When Deathmoon played MC game, he faced a math problem. When he found a ancient tomb and came in, he found two polynomials f(x) and g(x) no the wall, only did he calculate f(x) + g(x) correctly he could come in, can you help him?
For example:
f(x) = 2*x^5 + 3*x^3 + 7 + x^(-1),
g(x) = 3*x^4 + 2*x^3 + x - x^(-1).
Then,
f(x) + g(x) = 2*x^5 + 3*x^4 + 5*x^3 + x + 7
输入
Input end of EOF.
First I will give you two positive integers N and M(0 < N, M < 10) means the number of items in polynomials.
Then N lines follow, each line contains two integers c(-100 < c < 100) and p(-10 <= p <= 10), c means the coefficients and p means the power.
Then M lines follow, each line contains two integers c(-100 < c < 100) and p(-10 <= p <= 10), c means the coefficients and p means the power.
And for each test, every polynomial' s power is descending, it means pi > pj when i < j.
c != 0 && p != 0.
输出
You should output (f(x) + g(x))' s expression for coefficients and powers in descending.
样例输入
```
4 4
2 5
3 3
7 0
1 -1
3 4
2 3
1 1
-1 -1
```
样例输出
```
2 5
3 4
5 3
1 1
7 0
```
提示

f(x) = coefficients * x ^ power;
The sample input: 
f(x) = 2*x^5 + 3*x^3 + 7 + x^(-1),
g(x) = 3*x^4 + 2*x^3 + x - x^(-1).
The sample output: 
f(x) + g(x) = 2*x^5 + 3*x^4 + 5*x^3 + x + 7

题意：显而易见

思路：纯模拟，让下标确定为正数就是了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1000;

int n,m;
int arr[MAXN];
int vis[MAXN];

int main(){
	while (scanf("%d%d", &n, &m) != EOF){
		int c,p;
		memset(vis,0,sizeof(vis));
		for (int i = 0; i < n; i++){
			scanf("%d%d", &c, &p);
			arr[p+10] = c;
			vis[p+10] = 1;  //确保正数	
		}
		for (int i = 0; i < m; i++){
			scanf("%d%d", &c, &p);
			if (vis[p+10]){
				arr[p+10] += c;
				if (arr[p+10] == 0)
					vis[p+10] = 0;
			}
			else {
				vis[p+10] = 1;
				arr[p+10] = c;
			}
		}
		for (int i = 20; i >= 0; i--)
			if (vis[i])
				printf("%d %d\n", arr[i], i-10);
	}
	return 0;
}