CodeForces 716C Plus and Square Root(规律)

Description

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 t
4000
hat 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.

Examples

input
3
output
14
16
46
input
2
output
999999999999999998
44500000000
input
4
output
2
17
46
97

题目大意

初始时屏幕上显示为2，级别数level是1，然后进行x次+操作，每一次都将屏幕上的值+level，当屏幕上的数是一个平方数时，可以将其开平方，此时level+1,同时需要满足开万平方后的数是新的level的倍数。现在ZM 想要升到n+1级，问每一次需要操作的次数x是多少.

解题思路

根据前几步显示的值，我们可以总结出对于当前的级别level，其所需要进行的操作次数是，[level∗(level−1)]2−(level−1)∗(level−2)level−1=level∗level∗(level−1)−level+2[level∗(level−1)]2−(level−1)∗(level−2)level−1=level∗level∗(level−1)−level+2(level>=3),当level=2时输出2.

代码实现

#include<bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
typedef long long ll;
int main()
{
IO;
int n;
cin>>n;
cout<<"2"<<endl;
for(ll level=3;level<=n+1;level++)
cout<<level*level*(level-1)-level+2<<endl;
return 0;
}