A - Filling Diamonds_补全 暑假日常训练7.12
A - Filling Diamonds

题目描述

You have integer $n$. Calculate how many ways are there to fully cover belt-like area of 4n−2 triangles with diamond shapes.

Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.

2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.

Please look at pictures below for better understanding.

A - Filling Diamonds_#include_02

On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.

A - Filling Diamonds_#include_03

These are the figures of the area you want to fill for $n=1,2,3,4$

You have to answer t independent test cases.

输入

The first line contains a single integer t(1 ≤ t ≤ 10^4) — the number of test cases.

Each of the next $t$ lines contains a single integer n(1 ≤ n ≤ 10^9).

输出

For each test case, print the number of ways to fully cover belt-like area of 4n−2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^18.

样例

Input

2
2
1

Output

2
1

提示

In the first test case, there are the following 2 ways to fill the area:

A - Filling Diamonds_#include_04

In the second test case, there is a unique way to fill the area:

A - Filling Diamonds_#include_05

解析

当我们在自己凑方案的时候就可以发现

当我们在选定一个竖着的菱形作为第一个插入的菱形时，此菱形的左上角，左下角（右上角，右下角）就必须需要一个横着的菱形来补全

而当你补上一个菱形之后你会发现你需要一直补横着的菱形才能补全整个图形

所以对于任意的一个竖着的菱形，摆放的方案数就只有一个

所以当有n个竖着的菱形的时答案就为n

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+10;
int a[N];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		LL n;
		scanf("%lld",&n);
		printf("%lld\n",n);
	}
	return 0;
}