A - Print 341

思路:

给你一个整数N 有N个0和N + 1个1组成 0 1 交替输出1 

解法:

输出10最后输出最后剩余的1即可

Code:

#include <bits/stdc++.h>
    
using namespace std;
    
int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int N; cin >> N;
    for (int i = 1; i <= N; ++i) {
        cout << "10";
    }
    cout << "1\n";
    return 0;
}

Code:

N = int(input())
print("10" * N + "1")

B - Foreign Exchange

思路:

解法:

Code: