【Codeforces Round #372 (Div. 2)】Codeforces 716C Plus and Square Root

ZS the Coder is playing a game. There is a number displayed on the
screen and there are two buttons, ’ + ’ (plus) and ” (square root).
Initially, the number 2 is displayed on the screen. There are n + 1
levels in the game and ZS the Coder start at the level 1.

When ZS the Coder is at level k, he can :

Press the ' + ' button. This increases the number on the screen by exactly k. So, if the number on the screen was x, it becomes x + k.
Press the '' button. Let the number on the screen be x. After pressing this button, the number becomes . After that, ZS the Coder

levels up, so his current level becomes k + 1. This button can only be
pressed when x is a perfect square, i.e. x = m2 for some positive
integer m.

Additionally, after each move, if ZS the Coder is at level k, and the
number on the screen is m, then m must be a multiple of k. Note that
this condition is only checked after performing the press. For
example, if ZS the Coder is at level 4 and current number is 100, he
presses the ” button and the number turns into 10. Note that at this
moment, 10 is not divisible by 4, but this press is still valid,
because after it, ZS the Coder is at level 5, and 10 is divisible by
5.

ZS the Coder needs your help in beating the game — he wants to reach
level n + 1. In other words, he needs to press the ” button n times.
Help him determine the number of times he should press the ’ + ’
button before pressing the ” button at each level.

Please note that ZS the Coder wants to find just any sequence of
presses allowing him to reach level n + 1, but not necessarily a
sequence minimizing the number of presses. Input

The first and only line of the input contains a single integer n
(1 ≤ n ≤ 100 000), denoting that ZS the Coder wants to reach level
n + 1. Output

Print n non-negative integers, one per line. i-th of them should be
equal to the number of times that ZS the Coder needs to press the
’ + ’ button before pressing the ” button at level i.

Each number in the output should not exceed 1018. However, the number
on the screen can be greater than 1018.

It is guaranteed that at least one solution exists. If there are
multiple solutions, print any of them.

贪心地按’+’，每次加到k^2*(k+1)^2，然后开根号，再重复上述过程。
可以发现，每次加的次数就是i * (i+1) * (i+1)-i+1。

#include<cstdio>
#include<cstring>
#define LL long long
int main()
{
    int n;
    LL i;
    scanf("%d",&n);
    printf("2\n");
    for (i=2;i<=n;i++)
      printf("%I64d\n",i*(i+1)*(i+1)-i+1);
}