Save the problem!
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the
given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins
as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece.
Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations
of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee
there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Example
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
题意：给出方案数，求出满足这个方案数的任意一个面值搭配和硬币个数和价钱。
思路：只要是满足这个方案数的就可以所以就找1和2就可以了。那么如果是1和2的硬币，组合数是A种的话，那么它的价钱就是2*（A-1）(开始有点纠结这个怎么算的，想想发现：如果是x元由1元和2元分的话，方案数是（x/2）+1;,那么x就等于2*（A-1）;）,然后就只用特判1的情况。
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n==1)
{
printf("1 1
1
");
}
else
printf("%d 2
1 2
",2*(n-1));
}
}