pat 甲级 1009. Product of Polynomials (25)
1009. Product of Polynomials (25)
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 N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.
Output Specification:
For each test case you should output the product 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 up to 1 decimal place.
Sample Input
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output
3 3 3.6 2 6.0 1 1.6 思路:多项式相乘,模拟即可,要注意的是最终结果中系数为0的项不需要输出。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 30+5
#define M_MAX 2001
struct x {
int exp;
double coef = ;
bool vis = ;
};
x poly1[N_MAX],poly2[N_MAX];
x poly[M_MAX];
int n1, n2;
int main() {
cin >> n1;
for (int i = ; i < n1; i++)cin >> poly1[i].exp >> poly1[i].coef;
cin >> n2;
for (int i = ; i < n2; i++)cin >> poly2[i].exp >> poly2[i].coef;
for (int i = ; i < n1;i++) {
for (int j = ; j < n2;j++) {
int exp = poly1[i].exp + poly2[j].exp;
poly[exp].vis = ;
poly[exp].exp= exp;
poly[exp].coef+= poly1[i].coef*poly2[j].coef;
}
} int num = ;
//系数为0的项不用输出!!!!!!!!!
for (int i = M_MAX-; i >= ; i--) if (poly[i].vis&&poly[i].coef!=) num++;
cout << num << " ";
for (int i = M_MAX-; i >=;i--) {
if (poly[i].vis&&poly[i].coef != ) {
num--;
printf("%d %.1f%c",poly[i].exp,poly[i].coef,num==?'\n':' ');
}
}
return ;
}
pat 甲级 1009. Product of Polynomials (25)的更多相关文章
- PAT 甲级 1009 Product of Polynomials (25)(25 分)(坑比较多,a可能很大,a也有可能是负数,回头再看看)
1009 Product of Polynomials (25)(25 分) This time, you are supposed to find A*B where A and B are two ...
- PAT甲 1009. Product of Polynomials (25) 2016-09-09 23:02 96人阅读 评论(0) 收藏
1009. Product of 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, Yu ...
- PAT甲级——1009 Product of Polynomials
PATA1009 Product of Polynomials Output Specification: For each test case you should output the produ ...
- 【PAT】1009. Product of Polynomials (25)
题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1009 分析:简单题.相乘时指数相加,系数相乘即可,输出时按指数从高到低的顺序.注意点:多项式相 ...
- PAT Advanced 1009 Product of Polynomials (25 分)(vector删除元素用的是erase)
This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: Each ...
- PATA 1009. Product of Polynomials (25)
1009. Product of 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, Yu ...
- 1009 Product of Polynomials (25分) 多项式乘法
1009 Product of Polynomials (25分) This time, you are supposed to find A×B where A and B are two po ...
- PAT 1009 Product of Polynomials (25分) 指数做数组下标,系数做值
题目 This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: E ...
- 【PAT甲级】1009 Product of Polynomials (25 分)
题意: 给出两个多项式,计算两个多项式的积,并以指数从大到小输出多项式的指数个数,指数和系数. trick: 这道题数据未知,导致测试的时候发现不了问题所在. 用set统计非零项时,通过set.siz ...
随机推荐
- shell中字符串基本用法
前言 今天在写脚本时,发现前阶段使用过的一些用法还是需要去百度查找,并且找到的答案还需要自己去筛选,耽误写脚本时间,这一篇对字符串之间的比较和逻辑判断都非常详细,借鉴他人之作,资源共享. 本片主要说明 ...
- iView - Form中想要重置DatePicker生效,必须给DatePicker绑定value属性
Form中想要重置DatePicker生效,必须给DatePicker绑定value属性
- ubuntu16.04更换镜像源
1.备份原有 cp /etc/apt/sources.list /etc/apt/sources.list.old 2.打开阿里巴巴镜像源: https://opsx.alibaba.com/mir ...
- kubernetes搭建dashboard报错
warningconfigmaps is forbidden: User "system:serviceaccount:kube-system:kubernetes-dashboard&qu ...
- vue 组件的书写
简单的来说是 vue组件最核心的就是props和自定义函数,来实现组件的开发 最简单的一个组件 子组件如下: <template> <div class="bgClass& ...
- A1075 PAT Judge (25)(25 分)
A1075 PAT Judge (25)(25 分) The ranklist of PAT is generated from the status list, which shows the sc ...
- python单元测试用例
demo1.py #!/usr/bin/python # encoding: utf-8 def hello(): print "i am in demo1" def add(x, ...
- Android Kotlin适用小函数
都是一些Android适用的Kotlin小函数. 1.点击空白隐藏键盘 //点击空白隐藏键盘 override fun onTouchEvent(event: MotionEvent): Boolea ...
- 15,scrapy中selenium的应用
引入 在通过scrapy框架进行某些网站数据爬取的时候,往往会碰到页面动态数据加载的情况发生如果直接用scrapy对其url发请求,是获取不到那部分动态加载出来的数据值,但是通过观察会发现,通过浏览器 ...
- 1 django
1.MVC 大部分开发语言中都有MVC框架 MVC框架的核心思想是:解耦 降低各功能模块之间的耦合性,方便变更,更容易重构代码,最大程度上实现代码的重用 m表示model,主要用于对数据库层的封装 v ...