【PAT】A1002 A+B for Polynomials (25)-PAT甲级真题
A1002 A+B for Polynomials (25)
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
【polynomial:多项式的 nonzero term:非零项 decimal：十进制的
exponents:指数 exponential:指数的 coefficient:系数 respectively：分别地，依次地
】
1、自解（哈希表）
#include
#include
#include
using namespace std;
void AB_polynomial(vector& s1, vector& s2)
{
unordered_map m;
for (int i = 1; i <= s1[0]*2; i += 2)
{
m.insert(make_pair(s1[i], s1[i + 1]));
}
for (int j = 1; j <= s2[0]*2; j += 2)
{
if (m.count(s2[j])) { m[s2[j]] += s2[j + 1]; }
else{ m.insert(make_pair(s2[j], s2[j + 1])); }
}
int len = m.size();
cout << len << ' ';
for (auto it = m.begin();it!=m.end();it)
{
cout << it->first <<' ' << it->second;
it++;
if (it != m.end())cout << ' ';//最后没有多余的空格
}
}
int main()
{
vectors1,s2;
float a;
cin >> a; s1.push_back(a);
for (int i = 1; i <= s1[0]*2; i++)
{
cin >> a; s1.push_back(a);
}
cin >> a; s2.push_back(a);
for (int i = 1; i <= s2[0] * 2; i++)
{
cin >> a; s2.push_back(a);
}
/*s1 = {2,1,2.4,0,3.2};
s2 = { 2,2,1.5,1,0.5 };*/
AB_polynomial(s1, s2);
} Conclusion：
1、对于unordered_map，在插入键值对时要用：
m.insert(make_pair(a,b));2、查找键值：
m.count(a)!=0//表示m表中存在a为关键字的数，这个数是m[a]3、可以用迭代器遍历哈希表：
for(auto it = m.begin();it!=m.end();it++)
{
cout<4、对于vector的赋值方法有两种：
（1）直接赋值
vector s;
s = {1,2,3,4,5}; （2）cin函数接收元素赋值：
int a;
cin>>a;
s.push_back(a);//将a赋值给s2、柳神方法
直接用数组记录这些值....引发了我对hash和基础数组的区别的思考.......
优点：hash最大的方便在于可以快速查找元素，所以对于需要搜索大量数据的情况，hash的复杂度还是更低的！
缺点：需要额外创建一个hash表并且输入元素，费时间空间
总结：可以用hash但没那么必要！柳神yyds555
It is given that 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000.所以直接设置c1[1001]数组就好啦
#include
#include
using namespace std;
int main()
{
float c[1001] = { 0 };
int a; cin >> a;
int t; float num;
for (int i = 0; i < a; i++)
{
scanf_s("%d %f", &t, &num);
c[t] += num;
}
int b; cin >> b;
for (int j = 0; j < b; j++)
{
scanf_s("%d %f", &t, &num);
c[t] += num;
}
int count = 0;
for (int i = 0; i < 1001; i++)
{
if (c[i])count++;
}
cout << count;
for (int i = 1000; i >= 0; i--)
{
if (c[i] != 0.0)printf(" %d %.1f", i, c[i]);
}
} 【补充：找到了一个用hash的致命缺点：输出的结果具有随机性，不能排序！虽然题目里面也没有明确要求，但是和范例的output就是不一样......想要修改，但是在hash里面排序简直太违背它的魅力了TVT放弃啦 柳神的数组方法就ok】