C. Plus and Square Roottime limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 becomesx + k.
- Press the '
. 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 '
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 '
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.
InputThe 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.
OutputPrint 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 '
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.
Examplesinput3output14
16
46input2output999999999999999998
44500000000input4output2
17
46
97NoteIn the first sample case:
On the first level, ZS the Coder pressed the ' + ' button 14 times (and the number on screen is initially 2), so the number became2 + 14·1 = 16. Then, ZS the Coder pressed the '
.
After that, on the second level, ZS pressed the ' + ' button 16 times, so the number becomes 4 + 16·2 = 36. Then, ZS pressed the '
.
After that, on the third level, ZS pressed the ' + ' button 46 times, so the number becomes 6 + 46·3 = 144. Then, ZS pressed the '
.
Note that 12 is indeed divisible by 4, so ZS the Coder can reach level 4.
Also, note that pressing the ' + ' button 10 times on the third level before levelling up does not work, because the number becomes6 + 10·3 = 36, and when the '
In the second sample case:
On the first level, ZS the Coder pressed the ' + ' button 999999999999999998 times (and the number on screen is initially 2), so the number became 2 + 999999999999999998·1 = 1018. Then, ZS the Coder pressed the '
.
After that, on the second level, ZS pressed the ' + ' button 44500000000 times, so the number becomes109 + 44500000000·2 = 9·1010. Then, ZS pressed the '
.
Note that 300000 is a multiple of 3, so ZS the Coder can reach level 3.
题目链接：
http://codeforces.com/contest/715/problem/A
http://codeforces.com/contest/716/problem/C
题目大意：
一开始你有数2，处于level1，你可以选择+或者开方，选择加那么现在的数字将加上当前的level，
选择开方当且仅当你拥有的数十完全平方数，并且开方后level会加1，还要满足开方后所剩的数字是新的level的整数倍。
问从level1到level n+1 每一次需要+的次数【任意解即可】
题目思路：
【数学规律】
如果当前level=1，那么需要两次+，把数字变为4，开方后为2，还是2的倍数。
当level i(i>1)时，通过找规律可以知道，到达level i+1 需要的+的次数为i*(i+1)*(i+1)-(i-1)。
//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define N 50004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
LL n,m,lll,ans;
int main()
{
#ifndef ONLINE_JUDGEW
// freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
#endif
LL i,j,k;
// init();
// for(scanf("%d",&cass);cass;cass--)
// for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
// while(~scanf("%s",s))
while(~scanf("%I64d",&n))
{
for(i=;i<=n;i++)
{
if(i==)puts("");
else printf("%I64d\n",sqr(i+)*i-(i-));
}
}
return ;
}
/*
// //
*/