动态规划第五节课

#include <iostream>
#include <cstdio>
using namespace std;

long long dp[100010] = {0, 2, 7};

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 3; i <= n; i ++)
    {
        dp[i] = dp[i - 1] + 4 * i - 3;
    }
    printf("%lld\n", dp[n]);
    return 0;
}