Filling Diamonds
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You have integer nn. Calculate how many ways are there to fully cover belt-like area of 4n−24n−2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
22 coverings are different if some 22 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.
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.
These are the figures of the area you want to fill for n=1,2,3,4n=1,2,3,4.
You have to answer tt independent test cases.
Input
The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.
Each of the next tt lines contains a single integer nn (1≤n≤1091≤n≤109).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n−24n−2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10181018.
Example
input
代码语言：javascript复制2
2
1output
代码语言：javascript复制2
1Note
In the first test case, there are the following 22 ways to fill the area:
In the second test case, there is a unique way to fill the area:
思路：冷静下来，这是div2A题而已，一定是找规律的假题，我们观察到每一种新的方法唯一的不同之处就是竖放的菱形，就是n=1的那个图形，而n有n个不同位置的竖菱形，答案就是n
代码语言：javascript复制#include<bits/stdc .h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
x=0;char ch=getchar();ll f=1;
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=(x<<3) (x<<1) (ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res =sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x] =val;x =lb(x);}}//第x个加上val
ll t;
int main()
{
cin>>t;
for(int i=1;i<=t;i )
{
ll x;
cin>>x;
cout<<x<<endl;
}
return 0;
}
