This time, you are supposed to find A+B where A and B are two polynomials.

Input

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

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

这题翻译过来其实就是多项式加法

 #include<stdio.h>
#include<stdlib.h>
#include<math.h> int main()
{
int n;
int i,j;
double xi;
int zi;
double a[];
int cnt=; for( i=;i<; i++)
{
a[i] = ;
}
for( i=; i<; i++){
scanf("%d",&n);
for( j=; j<n; j++){
scanf("%d %lf",&zi,&xi);
a[zi] += xi;
}
} for( i=; i>=; i--){
if( fabs(a[i])>1e-)
cnt++;
}
printf("%d",cnt);
for( i=; i>=; i--){
if( fabs(a[i])>1e-)
printf(" %d %.1lf",i,a[i]);
} return ;
}

1002. A+B for Polynomials (25) (浮点数判0)的更多相关文章

  1. PAT 1002. A+B for Polynomials (25) 简单模拟

    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 ...

  2. PAT甲 1002. A+B for Polynomials (25) 2016-09-09 22:50 64人阅读 评论(0) 收藏

    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 ...

  3. 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 ...

  4. PAT (Advanced Level) Practice 1002 A+B for Polynomials (25 分) 凌宸1642

    PAT (Advanced Level) Practice 1002 A+B for Polynomials (25 分) 凌宸1642 题目描述: This time, you are suppos ...

  5. 1002 A+B for Polynomials (25)(25 point(s))

    problem 1002 A+B for Polynomials (25)(25 point(s)) This time, you are supposed to find A+B where A a ...

  6. 【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 ...

  7. 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 ...

  8. 1002. A+B for Polynomials (25)

    题目链接:https://www.patest.cn/contests/pat-a-practise/1002 原题如下: This time, you are supposed to find A+ ...

  9. 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

  1. Android偏好设置(2)为应用定义一个偏好设置xml

    1.Defining Preferences in XML Although you can instantiate new Preference objects at runtime, you sh ...

  2. DateFormat类

    package Format_daqo; import java.util.Date; import java.text.DateFormat; public class DateFormatTest ...

  3. C#控件置于底层或顶层

    btn.BringToFront();//置于顶层 btn.SendToBack();//置于底层

  4. 如何成为一名优秀的 iOS 开发工程师

    如果你是一位专业的iOS开发工程师,你应该为自己感到自豪.因为你能在强大的iOS系统下,一展身手实现自己和他人的想法,这是一件令人无比激动的事情. 作为一名iOS开发工程师,你一定想成为行业的佼佼者. ...

  5. Apache Tomcat 之路(二 部署web 应用程序)

    1.创建一个webapplication,不论是解压的应用程序包还是war包,在tomcat 上都能部署,这里提供一个简单的web项目:git地址:https://github.com/coderxi ...

  6. darknet+opencv在windows上的编译

    darknet 源码网站:https://github.com/pjreddie/darknet 技术支持官网:https://pjreddie.com/darknet/ darknet采用C++编写 ...

  7. 【C++】模板简述(五):类型萃取

    功能 类型萃取,在STL中用到的比较多,用于判断一个变量是否为POD类型. 简述来说可以用来判断出某个变量是内置类型还是自定义类型. 通过类型萃取,萃取到变量类型,对不同变量进行不同处理,可以提升程序 ...

  8. webpack2代码分割

    代码分割-CSS 要通过webpack打包CSS,像任何其他模块一样将CSS导入JavaScript代码,并使用css-loader(它输出CSS作为JS模块), 并可选地应用ExtractTextW ...

  9. vue HTTP请求(针对vue-resource)

    //初始化页面需要做什么事情 //点击后需要做什么事情 //鼠标.键盘.冒泡.默认行为等事件 //前端调用接口就是按需展示数据//created和mounted里面都可以做数据处理,唯一不同的是cre ...

  10. sql server 强制关闭连接

    USE master; GO DECLARE @SQL VARCHAR(MAX); SET @SQL='' SELECT @SQL=@SQL+'; KILL '+RTRIM(SPID) FROM ma ...