1002. A+B for Polynomials (25)

1002. A+B for Polynomials (25)

Question

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

这道题目的思路完全是按着人的直观想法来的。

用数组在表示一个多项式，每一个index表示次数，其中的value就是系数。所以要一个1001大小的、初始化为0的数组。

然后把第一个不等式的系数按照次数索引加进去，再把第二个不等式的系数按照次数索引加进去。

最后数一下数组中不为0的个数，就是新多项式的K，然后把数组倒着输出不为0的值即可啦！

最直观的代码

#include <iostream>

using namespace std;

const int MAX_EXP = 1001;

int main(int argc, const char * argv[]) {
double p1[MAX_EXP] = {0}, p2[MAX_EXP] = {0}, p3[MAX_EXP] = {0};
int K1, K2, exp;
double coefficient;

cin >> K1;
for (int i = 0; i < K1; ++i) {
cin >> exp >> coefficient;
p1[exp] = coefficient;
}

cin >> K2;
for (int i = 0; i < K2; ++i) {
cin >> exp >> coefficient;
p2[exp] = coefficient;
}

int K3 = 0;
for (int i = 0; i < MAX_EXP; ++i) {
p3[i] = p1[i] + p2[i];
if (p3[i] != 0) {
++K3;
}
}

cout << K3;

for (int i = MAX_EXP - 1; i >= 0; --i) {
if (p3[i] != 0) {
//这里一定要注意！输出的格式！保留1位小数！
printf(" %d %.1lf", i, p3[i]);
}
}

return 0;
}

稍微改进一点儿

#include <iostream>

using namespace std;

const int MAX_EXP = 1001;

int main(int argc, const char * argv[]) {
double p[MAX_EXP] = {0};
int exp = 0, K = 0 ;

cin >> K;
for (int i = 0; i < K; i++) {
cin >> exp;
cin >> p[exp];
}

cin >> K;
for (int i = 0; i < K ;i++) {
double temp = 0;
cin >> exp >> temp;
p[exp] += temp;
}

K = 0;
for (int i = 0; i < MAX_EXP; i++) {
if (p[i] != 0) {
K++;
}
}
cout << K;

for (int i = MAX_EXP - 1; i >= 0; i--) {
if (p[i] != 0) {
//这里一定要注意！输出的格式！保留1位小数！
printf(" %d %.1lf", i, p[i]);
}
}
return 0;
}

别人的解法

#include<iostream>
#include <vector>
#include <algorithm>
#include <iomanip>
using namespace std;

bool  compare(pair<int,float> a,pair<int,float> b) {
return a.first > b.first;
}

int main() {
int M,N;
vector<pair<int,float> >  vec;
vector<pair<int,float> >  vec1;
vector<pair<int,float> >::iterator it,it1;

float exp=0;
float coe=0;

cin>>M;
while(M--) {
cin>>exp>>coe;
vec.push_back(make_pair(exp,coe));
}

cin>>N;
while(N--) {
cin>>exp>>coe;
vec.push_back(make_pair(exp,coe));
}

sort(vec.begin(),vec.end(),compare);

for (it=vec.begin();it!=vec.end();it=it1){
for (it1=it+1; it1!=vec.end()
&& (*it).first == (*it1).first; it1++) {
(*it).second += (*it1).second;
}
if ((*it).second !=0) {
vec1.push_back(*it);
}
}

cout<<vec1.size();
for (it=vec1.begin();it!=vec1.end();it++) {
cout<<" "<<(*it).first
<<" "<<fixed<<setprecision(1)<<(*it).second;
}
cout<<endl;

return 0;
}