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PAT Advanced Level 1002 A+B for Polynomials

原题传送门 1. 问题描述 2. Solution 1、思路分析定义数组保存结果，依次读入A, B行，指数为下标，系数为值累加到结果数组中 2、代码实现 // PAT Advance Level 1002 // Ye Qiu #include <iostream> #include <string> using namespace std; /* 定义数组保存结果，依次读入A, B行

PAT 甲级题解

PAT 甲级 1001 A+B Format PAT 甲级 1002 A+B for Polynomials PAT 甲级 1003 Emergency DFS PAT 甲级 1004 Counting Leaves PAT 甲级 1005 Spell It Right 模拟字符串 PAT 甲级 1006 Sign In and Sign Out 字符串 PAT 甲级 1007 Maximum Subsequence Sum dp PAT 甲级 1008 Elev

PAT-1002 A+B for Polynomials

1002 A+B for Polynomials part 1, 1.1 自己解法(未全对) 使用两个map去存储数据 #include <iostream> using namespace std; #include <map> #include <algorithm> #include <iomanip> class MyCompare { public: bool operator()(int v1, int v2) {

1009 Product of Polynomials (25 分)（模拟）

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 i

PAT 甲级 1002 A+B for Polynomials

题目描述题目理解题目大意是给我们两个多项式序列，要求计算出两个多项式的和示例给出的answer格式是：项数X的次数值X的次数值X的次数值321.512.903.2 代码实现用两个数组记录输入；遍历一遍相加；把非0项统计并用ans记录；从大到小遍历数组从而实现X的次数从高到低；用printf控制

PAT-Advanced-【1002】A+B for Polynomials

【1002】 A+B for Polynomials (25 分) 1 //1002 A+B for Polynomials (25 分) 2 #include<iostream> 3 #include<iomanip> 4 using namespace std; 5 double a[1001] = { 0 }; 6 int main() { 7 int k,n,count=0; 8 double x; 9 int group

PAT-A1009 Product of Polynomials (多项式乘法)

A1009 Product of Polynomials (多项式乘法) 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 pol

[PAT] A1002 A+B for Polynomials (23分!)

A1002 A+B for Polynomials Input SpecificationOutput SpecificationSample InputSample OutputCode 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

1009 Product of 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 N1 aN1 N2 aN2 ... NK aNK where

1002 A+B for Polynomials（PAT，C语言）

目录题目代码测试结果题目 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 代码 #include <stdio.h> int main() { int i,j,m,x; float p[1001]={0},q; for(i=0;i<2;i++){ scanf("%d",&m); for(

Gauss型（Gaussian quadrature）求积公式和方法

目录 0、Gauss型积分通用形式 1、Gauss–Legendre quadrature勒让德 2、Gauss–Laguerre quadrature拉盖尔——积分区间[0，inf] 3、Chebyshev–Gauss quadrature切比雪夫 0、Gauss型积分通用形式 The integration problem can be expressed in a slightly more general way b

1002 A+B for Polynomials (25分)

注意系数为\(0\)的多项式要消去哦 map<int,double,greater<int>> mp; int n; int main() { for(int i=0;i<2;i++) { cin>>n; for(int j=0;j<n;j++) { int p; double a; scanf("%d%lf

PAT 甲级 1009 Product of Polynomials

https://blog.csdn.net/a845717607/article/details/86530433 map使用：声明：map<键，值> 变量名；存储： scanf("%d %lf",&e,&coe); mp1[e]=coe; 遍历：for(map<int,double>::iterator it=mp1.begin();it!=mp1.end();++it) 逆序遍历：for(map<int,doub

1002 A+B for Polynomials (25分)

1002 A+B for Polynomials (25分) #include<stdio.h> int main() { double a[1010]={0.0},b[1010]={0.0},c[1010]={0.0}; int n1,n2; scanf("%d",&n1); for(int i=0;i<n1;i++){ int a0; scanf("%d",&a0

1009 Product of Polynomials (25分)

1009 Product of 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

1002 A+B for Polynomials (25分)

1.多项式为0，直接输出0 2.系数为0的项不输出 3.注意double类型的判0 1 #include <iostream> 2 #include <vector> 3 #include <iomanip> 4 #include <cmath> 5 using namespace std; 6 7 int main() 8 { 9 vector<int> a, b;//次数 10 vec

1002 A+B for Polynomials

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

1009 Product of Polynomials

多项式加法——MOOC《零基础学Java语言》

多项式加法——MOOC《零基础学习Java语言》 #第五周编程题多项式加法注意： 1.多项式全为0，显示0。 2.最高幂次前面不显示加减符号。 3.幂次为1，不显示. 4.系数为1或-1，不显示。 import java.util.Scanner; public class Main { public static void main(String[] args) {

1009 Product of Polynomials

题目描述 This time, you are supposed to find A×B where A and B are two polynomials. 输入格式 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 w

1002 A+B for Polynomials

题目链接 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: where K is the number o

PAT甲级 1002 A+B for Polynomials

PAT甲级 1002 A+B for Polynomials 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:

PAT-A1002 A+B for Polynomials (25分)

原题 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 aN

A1002. A+B for Polynomials

题目描述　　This time, you are supposed to find A+B where A and B are two polynomials 输入格式　　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

入门模拟A1009 Product of Polynomials(25)

2019-12-23 14:54:40 #include <bits/stdc++.h> #include<math.h> using namespace std; const int MAX_LEN = 2004; int main(){ double temp1[MAX_LEN]={0}; double temp2[MAX_LEN]={0}; int tem1; cin>>tem1; for(int i =0;i&