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, Yue

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

  1. 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 ...

  2. 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 ...

  3. PAT甲级——1009 Product of Polynomials

    PATA1009 Product of Polynomials Output Specification: For each test case you should output the produ ...

  4. 【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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 【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

  1. BZOJ2118: 墨墨的等式(最短路 数论)

    题意 墨墨突然对等式很感兴趣,他正在研究a1x1+a2y2+…+anxn=B存在非负整数解的条件,他要求你编写一个程序,给定N.{an}.以及B的取值范围,求出有多少B可以使等式存在非负整数解. So ...

  2. 七、MySQL 选择数据库

    MySQL 选择数据库 在你连接到 MySQL 数据库后,可能有多个可以操作的数据库,所以你需要选择你要操作的数据库. 从命令提示窗口中选择MySQL数据库 在 mysql> 提示窗口中可以很简 ...

  3. ngin负载均衡集群(一)

    一.nginx负载均衡集群介绍: 1.反向代理与负载均衡概念简介严格地说, nginx仅仅是作为 Nginx Proxy反向代理使用的,因为这个反向代理功能表现的效果是负载均衡集群的效果,所以本文称之 ...

  4. 17.VUE学习之- v-for指令的使用方法

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  5. php实现的三个常用加密解密功能函数示例

    目录 算法一: 算法二: 算法三(改进第一个加密之后的算法) 本文实例讲述了php实现的三个常用加密解密功能函数.分享给大家供大家参考,具体如下: 算法一: //加密函数 function lock_ ...

  6. linux 的安装

    3linux 软件安装 3.1 vm ware 软件安装 双击VMware-workstation-full-10.0.2-1744117.1398244508.exe 单击下一步 单击下一步 选择典 ...

  7. python面向对象之反射和内置方法

    一.静态方法(staticmethod)和类方法(classmethod) 类方法:有个默认参数cls,并且可以直接用类名去调用,可以与类属性交互(也就是可以使用类属性) 静态方法:让类里的方法直接被 ...

  8. Python猫荐书系列之七:Python入门书籍有哪些?

    本文原创并首发于公众号[Python猫],未经授权,请勿转载. 原文地址:https://mp.weixin.qq.com/s/ArN-6mLPzPT8Zoq0Na_tsg 最近,猫哥的 Python ...

  9. #3 working with data stored in files && securing your application

    This chapter reveals that you can use files and databases together to build PHP application that waa ...

  10. Python中*和**的区别

    Python中,(*)会把接收到的参数形成一个元组,而(**)则会把接收到的参数存入一个字典 我们可以看到,foo方法可以接收任意长度的参数,并把它们存入一个元组中 >>> def ...