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 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 结尾无空行Sample Output:
3 2 1.5 1 2.9 0 3.2 结尾无空行
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node
{
int z;
float x;
};
bool compare(node A, node B)
{
return A.z > B.z;
}
int main()
{
vector<node> A, B, C;
node tmp;
int a, b;
cin >> a;
for (int i = 0; i < a; i++)
{
cin >> tmp.z >> tmp.x;
A.push_back(tmp);
}
cin >> b;
for (int i = 0; i < b; i++)
{
cin >> tmp.z >> tmp.x;
B.push_back(tmp);
}
// 排序
sort(A.begin(), A.end(), compare);
sort(B.begin(), B.end(), compare);
// 双指针法
int i, j;
for (i = 0, j = 0; i < a && j < b;)
{
if (A[i].z > B[j].z)
{
tmp = A[i++];
C.push_back(tmp);
}
else if (A[i].z < B[j].z)
{
tmp = B[j++];
C.push_back(tmp);
}
else
{
tmp.z = A[i].z;
tmp.x = A[i++].x + B[j++].x;
// 系数为零就抵消了
if (tmp.x)
{
C.push_back(tmp);
}
}
}
// 余值按序插入，2个while只有一个会被执行
while (i < a)
{
tmp = A[i++];
C.push_back(tmp);
}
while (j < b)
{
tmp = B[j++];
C.push_back(tmp);
}
// 输出
cout << C.size();
for (node it : C)
{
printf(" %d %.1f", it.z, it.x);
}
system("pause");
return 0;
}
受最近数据结构的影响，思维还没转变过来。想到的是传统的结构体解法，合并多项式借用了归并排序的思想，先将2个多项式降序排序，然后双指针法合并。卡壳的主要原因是没有考虑多项式相加为零要剔除的情况。
#include <iostream>
using namespace std;
int main()
{
double N[1005] = {
0};
for (int i = 0; i < 2; i++)
{
int len;
cin >> len;
for (int k = 0; k < len; k++)
{
int z;
double x;
cin >> z >> x;
N[z] += x;
}
}
int count = 0;
for (int i = 0; i <= 1000; i++)
{
if (N[i] != 0)
{
count++;
}
}
cout << count;
for (int i = 1000; i >= 0; i--)
{
if (N[i] != 0)
{
printf(" %d %.1f", i, N[i]);
}
}
return 0;
}
数组下标表示指数，数组元素表示系数。简单明了。