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:
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 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
分析：将系数相同的项，变量相加，就ok了，但是注意一个地方，系数可以为0，但是变量不可以为0（哭了，不知道有这种规定），所以输出时把变量为0的剔除掉
代码如下：
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 #include <unordered_map> 5 using namespace std; 6 7 struct term { 8 int times; 9 double num; 10 }; 11 12 bool cmp(term a, term b) { return a.times > b.times; } 13 14 unordered_map<int, int> mapp; 15 vector<term> poly; 16 17 int main() 18 { 19 int a, b, exp; 20 double coe; 21 scanf("%d", &a); 22 for (int i = 0; i < a; i++) 23 { 24 scanf("%d %lf", &exp, &coe); 25 auto it = mapp.find(exp); 26 if (it != mapp.end()) 27 { 28 //在map中找到 29 poly[it->second].num += coe; 30 } 31 else 32 { 33 //没找到 34 int count = poly.size(); 35 mapp[exp] = count; //map中存入该键值对 36 term temp = { exp,coe }; 37 poly.push_back(temp); 38 } 39 } 40 scanf("%d", &b); 41 for (int i = 0; i < b; i++) 42 { 43 scanf("%d %lf", &exp, &coe); 44 auto it = mapp.find(exp); 45 if (it != mapp.end()) 46 { 47 //在map中找到 48 poly[it->second].num += coe; 49 } 50 else 51 { 52 //没找到 53 int count = poly.size(); 54 mapp[exp] = count; //map中存入该键值对 55 term temp = { exp,coe }; 56 poly.push_back(temp); 57 } 58 } 59 sort(poly.begin(), poly.end(), cmp); 60 int count=0; 61 for (auto it = poly.begin(); it != poly.end(); it++) 62 { 63 if (it->num != 0) count++; 64 } 65 cout << count; 66 for (auto it = poly.begin(); it != poly.end(); it++) 67 { 68 if(it->num!=0) 69 printf(" %d %.1lf", it->times, it->num); 70 } 71 return 0; 72 }
这里用了一个vector和一个 map，因为一开始看到这个想到的就是用数组，而寻找多项式的各个项对应的位置又想到了用字典。
看有人用的是map而不是unoder_map，然后利用map的自动键值从小到大排序来实现系数的有序。我使用数组来记录多项式，而他的做法用字典的键值对来记录多项式。
这里贴上他的做法，我觉得非常ok。
1 #include<cstdio> 2 #include<map> 3 using namespace::std; 4 map<int,double> s; 5 int main() 6 { 7 int i1,t1; 8 double t2; 9 for(int z=0;z<2;z++) 10 { 11 scanf("%d",&i1); 12 for(int i=0;i<i1;i++) 13 { 14 scanf("%d%lf",&t1,&t2); 15 if(s.count(t1)==0) 16 s[t1]=t2; 17 else 18 s[t1]+=t2; 19 } 20 } 21 i1=0; 22 for(map<int,double>::const_iterator m_it=s.begin();m_it!=s.end();m_it++) 23 { 24 if(m_it->second!=0.0&&m_it->second!=-0.0) 25 i1++; 26 } 27 printf("%d",i1); 28 for(map<int,double>::reverse_iterator m_it=s.rbegin();m_it!=s.rend();m_it++) 29 { 30 if(m_it->second!=0.0&&m_it->second!=-0.0) 31 printf(" %d %.1lf",m_it->first,m_it->second); 32 } 33 printf("\n"); 34 return 0; 35 } 36 37 --------------------- 38 作者：漂流瓶jz 39 来源：CSDN 40 原文：https://blog.csdn.net/qq278672818/article/details/54563738 41 版权声明：本文为博主原创文章，转载请附上博文链接！
