PAT 甲级 1002 A+B for Polynomials
https://pintia.cn/problem-sets/994805342720868352/problems/994805526272000000
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 (,) are the exponents and coefficients, respectively. It is given that 1,0.
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
代码:
#include <bits/stdc++.h>
using namespace std; double a[1010], b[1010];
int A, B;
double sum[1010]; int main() {
scanf("%d", &A);
for(int i = 1; i <= A; i ++) {
int x;
scanf("%d", &x);
scanf("%lf", &a[x]);
}
scanf("%d", &B);
for(int i = 1; i <= B; i ++) {
int y;
scanf("%d", &y);
scanf("%lf", &b[y]);
} int cnt = 0;
for(int i = 0; i <= 1000; i ++) {
if(a[i] && b[i])
sum[i] = a[i] + b[i];
else if(a[i])
sum[i] = a[i];
else if(b[i])
sum[i] = b[i];
}
for(int i = 0; i <= 1000; i ++) {
if(sum[i])
cnt ++;
}
printf("%d", cnt);
for(int i = 1000; i >= 0; i --) {
if(sum[i])
printf(" %d %.1lf", i, sum[i]);
}
printf("\n");
return 0;
}
PAT 甲级 1002 A+B for Polynomials的更多相关文章
- PAT 甲级 1002 A+B for Polynomials (25 分)
1002 A+B for Polynomials (25 分) This time, you are supposed to find A+B where A and B are two polyno ...
- PAT甲级 1002 A+B for Polynomials (25)(25 分)
1002 A+B for Polynomials (25)(25 分) This time, you are supposed to find A+B where A and B are two po ...
- PAT 甲级1002 A+B for Polynomials (25)
1002. A+B for 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 T ...
- PAT甲级——1002 A+B for Polynomials
PATA1002 A+B for Polynomials This time, you are supposed to find A+B where A and B are two polynomia ...
- pat甲级1002
1002. A+B for 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 T ...
- 【PAT】1002. A+B for Polynomials (25)
1002. A+B for Polynomials (25) This time, you are supposed to find A+B where A and B are two polynom ...
- PAT——甲级1009:Product of Polynomials;乙级1041:考试座位号;乙级1004:成绩排名
题目 1009 Product of Polynomials (25 point(s)) This time, you are supposed to find A×B where A and B a ...
- 甲级1002 A+B for Polynomials (25)
题目描述: This time, you are supposed to find A+B where A and B are two polynomials. Input Each input fi ...
- PAT Advanced 1002 A+B for Polynomials (25 分)(隐藏条件,多项式的系数不能为0)
This time, you are supposed to find A+B where A and B are two polynomials. Input Specification: Each ...
随机推荐
- MySql基本数据类型及约束
1. 常用的数据类型(data_type) 字符串类型 CHAR(n) : 固定长度 VARCHAR(n) : 可变长度 NCHAR(n) : 使用utf8存储,固定长度 NVARCHAR(n) : ...
- Maximum Subsequence Sum
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to ...
- ABAP术语-Customer Enhancement
Customer Enhancement 原文:http://www.cnblogs.com/qiangsheng/archive/2008/01/18/1043874.html Adjustment ...
- ABAP术语-Business Components
Business Components 原文:http://www.cnblogs.com/qiangsheng/archive/2007/12/26/1015254.html Group of re ...
- json_decode结果为null的几种原因
值只能是UTF-8编码,元素最后不能有逗号,元素不能使用单引号,元素值中间不能有空格和n.
- ReentrantLock详解
ReentrantLock概述 ReentrantLock是Lock接口的实现类,可以手动的对某一段进行加锁.ReentrantLock可重入锁,具有可重入性,并且支持可中断锁.其内部对锁的控制有两种 ...
- JAVA Web 项目中用到的技术
JSPServletTomcatMySQL MavenSpringMVCHibernatejQueryBootstrapAngularJSBootStrap Table 下边两个是移动APP开发要用到 ...
- ruby 比较符号==, ===, eql?, equal?
“==” 最常见的相等性判断 “==” 使用最频繁,它通常用于对象的值相等性(语义相等)判断,在 Object 的方法定义中,“==” 比较两个对象的 object_id 是否一致,通常子类都会重写覆 ...
- Linux命令备忘录:mount用于加载文件系统到指定的加载点
mount命令用于加载文件系统到指定的加载点.此命令的最常用于挂载cdrom,使我们可以访问cdrom中的数据,因为你将光盘插入cdrom中,Linux并不会自动挂载,必须使用Linux mount命 ...
- springmvc springboot 跨域问题(CORS)
官方文档:http://docs.spring.io/spring/docs/current/spring-framework-reference/html/cors.html springmvc s ...