PAT 1002. A+B for Polynomials (25)（多项式加法）（待修改）

题目

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, YueThis time, you are supposed to find A+B where A and B are two polynomials.InputEach 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.OutputFor 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 Input2 1 2.4 0 3.22 2 1.5 1 0.5Sample Output3 2 1.5 1 2.9 0 3.2

问题代码

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<iomanip>using namespace std;bool cmp(int a,int b){    return a>b;}int main(int argc, char *argv[]){    ios_base::sync_with_stdio(false);    double num[1001]={0};    bool visited[1001] = {false};    int k1;    cin >> k1;    int base;double eco;    vector<int> v;    while (k1--) {        cin >> base >> eco;        num[base] += eco;        if (!visited[base]) {            v.push_back(base);            visited[base] = true;        }    }    cin >> k1;    while (k1--) {        cin >> base >> eco;        num[base] += eco;        if (!visited[base]) {            v.push_back(base);            visited[base] = true;        }    }    cout<< v.size();    sort(v.begin(),v.end(),cmp);    for (int i = 0; i < v.size(); ++i) {        cout<<" "<< v[i] << " " << fixed << setprecision(1) <<num[v[i]];    }    cout << endl;    return 0;}