题解 | #公共子串计算#

参考了第一个高赞老哥的思路

#include <iostream>
using namespace std;

int main (){
    int n, first_number;  //n就是m, first_number是数组首项
    while(cin >> n){
        first_number = n*n - n + 1;
        for(int i = 1; i <= n; i++){
            cout << first_number;
            if(i < n) cout<<"+";
            first_number += 2;
        }
        cout << endl;
    }
    return 0;
}