PAT1002. A+B for Polynomials (25)

1002. A+B for 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, Yue

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

这个题注意两个地方， 第一：最后输出的总项数是针对多项式求和后每一项系数不为0的项。第二：求和后系数为零的项不输出。

#include<iostream>
#include<algorithm>
#include<iomanip>
#include<string>

typedef struct number
{
int z;
double x;
bool flag;
}num;

bool cmp(num a, num b)
{
return a.z>b.z;
}

int main()
{
std::ios::sync_with_stdio(false);
int n, m;
std::cin>>n;
num* p = new num
;
for(int i=0;i<n;++i)
{
std::cin>>p[i].z>>p[i].x;
p[i].flag = false;
}
std::cin>>m;
num* q = new num [m];
for(int i=0;i<m;++i)
{
std::cin>>q[i].z>>q[i].x;
q[i].flag = false;
}
num t[1001];
int dex=0;
for(int i=0;i<n;++i)
{
for(int j=0;j<m;++j)
{
if(p[i].z == q[j].z)
{
t[dex] = p[i];
t[dex].x += q[j].x;
q[j].flag = true;
p[i].flag = true;
}
}
if(p[i].flag == false)
{
t[dex] = p[i];
p[i].flag = true;
}
dex++;
}
for(int i=0;i<m;++i)
{
if(q[i].flag == false)
{
t[dex++] = q[i];
}
}
std::sort(t, t+dex, cmp);
int _dex=0;
for(int i=0;i<dex;++i)
if(t[i].x!=0)
_dex++;
std::cout<<_dex;
for(int i=0;i<dex;++i)
if(t[i].x!=0)
i==0?std::cout<<" "<<t[i].z<<" "<<std::fixed<<std::setprecision(1)<<t[i].x:std::cout<<" "<<t[i].z<<" "<<std::fixed<<std::setprecision(1)<<t[i].x;
std::cout<<std::endl;
return 0;
}