PTA 乙级 1010 一元多项式求导 (25分)

 1 #include<iostream>
 2 #include<vector>
 3 using namespace std;
 4 
 5 int main() {
 6     int n = 0;
 7     vector<int> num;    //vector动态处理，先不定义大小
 8     do {
 9         cin >> n;
10         num.push_back(n);
11     } while (cin.get() != ‘\n‘);
12     if (num[1] == 0)    //判断零多项式的情况
13         cout << "0 0";
14     else
15         cout << num[1] * num[0] << " " << num[1] - 1;
16     for (int i = 2; i + 1 <= (int)num.size(); i += 2) {        //以两个数为单元处理
17         if (num[i + 1] != 0) 
18             cout << " " << num[i + 1] * num[i] << " " << num[i + 1] - 1;
19     }
20     return 0;
21 }