纯粹的逻辑判断题

没有什么大的难点，但要注意以下几个坑:

1.系数为 0 的项不输出.

2.如果多项式 n 次项系数为正，则多项式开头不出 + 号.

3.如果一个高于 0 次的项，其系数的绝对值为 1，则无需输出 1.

4.如果 x 的指数为 0，则仅需输出系数即可.

逻辑判断主要集中在这些方面。
下面是代码部分，欢迎指正:

#include <bits/stdc++.h>
using namespace std;

signed main() {
    int n;
    cin >> n;
    for (int i = 0;i < n + 1;i ++) { //一个n项式，最多有n+1项
        int each;
        cin >> each;
        if (i == 0) {
            if (each != 0) { //判断最高次项系数，注意当为正整数时开头无符号
                if (each != 1 && each != -1) {
                    cout << each << "x^" << n;
                }
                else if (each == 1) cout << "x^" << n; //分开处理正负两种情况，下同
                else cout << "-x^" << n; 
            }
        }
        else if (i == n) { //判断常数项，注意结尾无"x^"
            if (each > 0) cout << "+" << each;
            else if (each < 0) cout << each;
        }
        else if (n - i == 1) { //判断一次项，注意结尾写为"x"
            if (each != 1 && each != -1) {
                if (each > 0) cout << "+" << each << "x";
                else if (each < 0) cout << each << "x";
            }
            else { //判断其他项
                if (each > 0) cout << "+x";
                else if (each < 0) cout << "-x";
            }
        }
        else {
            if (each != 1 && each != -1) {
                if (each > 0) cout << "+" << each << "x^" << n - i;
                else if (each < 0) cout << each << "x^" << n - i;
            }
            else {
                if (each > 0) cout << "+x^" << n - i;
                else if (each < 0) cout << "-x^" << n - i;
            }
        }
    }
}

给个赞吧谢谢！