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​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

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

题目很简单,说一下可能的wa点:
1. 末尾不要有空格。
2. 两项相加可能为零,此时要删除此项。
代码如下:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef struct node *ptonext;
typedef ptonext ploy, list;
struct node{
float n;
int e;
ptonext next;
ptonext pro;
};
list A, B;
list res;
ptonext A_head, A_tail;
ptonext B_head, B_tail;
ptonext r_head, r_tail;
int cnt;
list init()
{
ptonext T;
T = (list)malloc(sizeof(list));
T->n = ;
T->e = ;
T->pro = NULL;
T->next = NULL;
return T;
}
void caculate()
{
ptonext a, b, r;
ptonext T;
a = A_head;
b = B_head;
r = r_head;
a = a->next;
b = b->next;
while(a != A_tail && b != B_tail)
{
if(a->e == b->e){
T = (ptonext)malloc(sizeof(ptonext));
T->e = a->e;
T->n = a->n + b->n;
if(T->n == ){
a = a->next;
b = b->next;
continue;
}
T->pro = r;
r->next = T;
r = r->next;
a = a->next;
b = b->next;
cnt++;
}
else if(a->e > b->e){
T = (ptonext)malloc(sizeof(ptonext));
T->e = a->e;
T->n = a->n;
r->next = T;
T->pro = r;
r = r->next;
a = a->next;
cnt++;
}
else{
T = (ptonext)malloc(sizeof(ptonext));
T->e = b->e;
T->n = b->n;
r->next = T;
T->pro = r;
r = r->next;
b = b->next;
cnt++;
}
}
while(a != A_tail){
T = (ptonext)malloc(sizeof(ptonext));
T->e = a->e;
T->n = a->n;
r->next = T;
T->pro = r;
r = r->next;
a = a->next;
cnt++;
}
while(b != B_tail){
T = (ptonext)malloc(sizeof(ptonext));
T->e = b->e;
T->n = b->n;
r->next = T;
T->pro = r;
r = r->next;
b = b->next;
cnt++;
}
r->next = r_tail;
r_tail->pro = r;
}
int main()
{
int k;
float n;
int e;
A = init();
B = init();
res = init();
A_tail = init();
B_tail = init();
r_tail = init();
A_head = A;
B_head = B;
r_head = res;
cnt = ;
scanf("%d", &k);
while(k--){
scanf("%d %f", &e, &n);
ptonext T;
T = (ptonext)malloc(sizeof(ptonext));
T->n = n;
T->e = e;
A->next = T;
T->pro = A;
A = A->next;
}
A->next = A_tail;
A_tail->pro = A;
scanf("%d", &k);
while(k--){
scanf("%d %f", &e, &n);
ptonext T;
T = (ptonext)malloc(sizeof(ptonext));
T->n = n;
T->e = e;
B->next = T;
T->pro = B;
B = B->next;
}
B->next = B_tail;
B_tail->pro = B;
caculate();
printf("%d", cnt);
while(res->next != r_tail){
printf(" %d %.1f", res->next->e, res->next->n);
res = res->next;
}
printf("\n");
// system("pause");
return ;
}

1002 A+B for Polynomials (PAT (Advanced Level) Practice)的更多相关文章

  1. 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 ...

  2. PAT (Advanced Level) Practice(更新中)

    Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性 ...

  3. PAT (Advanced Level) Practice 1001-1005

    PAT (Advanced Level) Practice 1001-1005 PAT 计算机程序设计能力考试 甲级 练习题 题库:PTA拼题A官网 背景 这是浙大背景的一个计算机考试 刷刷题练练手 ...

  4. PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge someti ...

  5. PAT (Advanced Level) Practice 1027 Colors in Mars (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1027 Colors in Mars (20 分) 凌宸1642 题目描述: People in Mars represent the c ...

  6. PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1011 World Cup Betting (20 分) 凌宸1642 题目描述: With the 2010 FIFA World Cu ...

  7. PAT (Advanced Level) Practice 1046 Shortest Distance (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1046 Shortest Distance (20 分) 凌宸1642 题目描述: The task is really simple: ...

  8. PAT (Advanced Level) Practice 1042 Shuffling Machine (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1042 Shuffling Machine (20 分) 凌宸1642 题目描述: Shuffling is a procedure us ...

  9. PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642 题目描述: Being unique is so important to peo ...

随机推荐

  1. zabbix如何添加主机监控

    1,首先,监控的主机安装zabbix客户端.zabbix提供多种监控方式,我们这里监控的主机上边安装agentd守护端进行数据收集并监测. 其中客户端安装我们这里就不介绍了,请参考之前教程里边的客户端 ...

  2. leetcode 258. Add Digits——我擦,这种要你O(1)时间搞定的必然是观察规律,总结一个公式哇

    Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. ...

  3. SAM入门

    学了两天,会了点皮毛,这里只放代码. P3804 #include<iostream> #include<cstdio> #include<cmath> #incl ...

  4. python cmd 启动python项目报错:no module named “xxx”

    场景:使用pycharm编辑器启动pyhon项目时可以启动,但使用cmd启动时,会报:no module named “xxx”的错误,此时,有两种情况: 1.no module named “xxx ...

  5. JAVA线程同步 (二)notify()与notifyAll()-***

    编写多线程程序需要进行线程协作,前面介绍的利用互斥来防止线程竞速是来解决线程协作的衍生危害的.编写线程协作程序的关键是解决线程之间的协调问题,在这些任务中,某些可以并行执行,但是某些步骤需要所有的任务 ...

  6. c#.net常用函数列表

    .DateTime 数字型 System.DateTime currentTime=new System.DateTime(); 1.1 取当前年月日时分秒 currentTime=System.Da ...

  7. json和Jsonp 使用总结(2)

    1.Jsonp的使用 var phoneAgent = navigator.userAgent; var urlDomaintest = " "; function getHref ...

  8. 数据结构之动态顺序表(C实现)

    线性表有2种,分为顺序表和链表. 顺序表: 采用顺序存储方式,在一组地址连续的存储空间上存储数据元素的线性表(长度固定) 链表: 有3种,单链表.双向链表.循环链表(长度不固定) seqList.h ...

  9. ACM_递推题目系列之一涂色问题(递推dp)

    递推题目系列之一涂色问题 Time Limit: 2000/1000ms (Java/Others) Problem Description: 有排成一行的n个方格,用红(Red).粉(Pink).绿 ...

  10. Android 性能优化(23)*性能工具之「Heap Viewer, Memory Monitor, Allocation Tracker」Memory Profilers

    Memory Profilers In this document Memory Monitor Heap Viewer Allocation Tracker You should also read ...