浙大PAT甲级-1002

这道题是大一的时候就做过的，当时刚学运算符重载。现在我用的是链表来做的，也算是复习一下快忘光的数据结构吧。

#include <iostream>
#include<iomanip>
using namespace std;//节点（多项式的项）
struct listNode{int exp;double coe;listNode *link;listNode(){}listNode(int e, double c){exp = e;coe = c;link = NULL;}
};class list{
public:listNode *first;listNode *last;   //方便后面直接用就没设为private啦list(){first = NULL;last = NULL;}//输入，初始化链表void init(int n){int e;double c;for (int i = 0; i<n; i++){cin >> e >> c;insert(e, c);}}//插入节点void insert(int e, double c){listNode *newNode = new listNode(e,c);if (first == NULL){newNode->link = first;first =  newNode;last = newNode;}else{last->link = newNode;last = newNode;}last->link = NULL;}//将系数为0的节点都剪掉void merge(){while (first!=NULL && first->coe == 0){first = first->link;}if (first != NULL){listNode *m = first->link;listNode *n = first;while (m != NULL){if (m->coe == 0){n->link = m->link;}else{n = n->link;}m = m->link;}}}//链表长度int Length(){listNode *p = first;int cnt = 0;while (p != NULL){p = p->link;cnt++;}return cnt;}
};
int main()
{int p, q;while (cin >> p){list a, b, c;a.init(p);cin >> q;b.init(q);//多项式加法listNode *ca = a.first;listNode *cb = b.first;while (ca != NULL&&cb != NULL){if (ca->exp > cb->exp){c.insert(ca->exp, ca->coe);ca = ca->link;}else if (ca->exp == cb->exp){c.insert(ca->exp, ca->coe + cb->coe);ca = ca->link;cb = cb->link;}else{c.insert(cb->exp, cb->coe);cb = cb->link;}}if (ca != NULL){while (ca != NULL){c.insert(ca->exp, ca->coe);ca = ca->link;}}else if (cb != NULL){while (cb != NULL){c.insert(cb->exp, cb->coe);cb = cb->link;}}c.merge();cout << c.Length();listNode *m = c.first;while (m != NULL){cout << " " << m->exp << " " << setiosflags(ios::fixed)<< setprecision(1) << m->coe;//输出是有精度要求的哦m = m->link;}cout << endl;}return 0;
}

浙大PAT甲级-1002相关推荐

Pat甲级 1002 A+B for Polynomials
Pat甲级 1002 A+B for Polynomials 思路代码题目网址 https://pintia.cn/problem-sets/994805342720868352/problems ...
PAT甲级1002 A+B for Polynomials：[C++题解]字符串、多项式加法或高精度加法
文章目录题目分析题目链接题目分析本题数据范围是1000 ,可以开一个数组 X [ ] ,X[i ]表示多项式中次幂是i的系数是多少. 因此,本题可以开两个数组,对应相加,然后输出非零项即可. ...
浙大PAT甲级1040
浙大PAT甲级1040 原题问题分析原题 1040 Longest Symmetric String (25 分) Given a string, you are supposed to outp ...
PAT甲级 1002
PAT 甲级 1002 题目描述:多项式求和问题输入两个多项式,输出和的多项式(多项式按指数递减排列) Sample Input: 2 1 2.4 0 3.2 2 2 1.5 1 0.5 Sampl ...
浙大PAT甲级1019. General Palindromic Number (20)
1019. General Palindromic Number (20) 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 ...
PAT甲级 -- 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 ...
浙大PAT甲级1027. Colors in Mars (20)
1027. Colors in Mars (20) 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 People ...
浙大PAT甲级1006
1006. Sign In and Sign Out (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 ...
浙大 PAT 甲级1009
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 ...
PAT 甲级 1002
多项式加法两处需要注意的: 相加后的多项式的系数如果是0,就不能输出注意C++输出的格式 #include <bits/stdc++.h> using namespace std;st ...

浙大PAT甲级-1002

浙大PAT甲级-1002相关推荐

最新文章

热门文章