Problem Description
Give you a cube with a side length of n-1. Find the number of equilateral triangles with three points on the cube point. Each side must be parallel to a certain surface of Oxy, Oxz, Oyz. Now you need to count how many such triangles there are.Each point can only be on the boundary or inner point of the cube, and the three coordinates x, y, and z of each point must be integers.
Input
The first line contains an integer T(T<=1e5) . Then T test cases follow.
Each test case contains a single Integer n(0<=n<=1e18).
If n=0, output 0
Output
For each case, print an integer, which is the answer modulo 109+7
Sample Input
2
1
2
Sample Output
0
8
觀察發現三個點所有坐標都是整數的三角形只可能是第二個樣例的那八種情況，即類似(0, 0, 0), (1, 0, 1), (0, 1, 1)。證明的話正三角形一條邊平行於棱的情況很容易發現是不成立的，其他情況emmm
這樣就只需要知道邊長為n - 1的立方體包含多少邊長為1，2...n-1的立方體，再將答案乘以8即可。
\(\Sigma_{i = 1}^{n-1}8\times i^3 =2(n(n-1))^2\)。n-1必須要先模一下模數！
#include <bits/stdc++.h>
#define mod 1000000007
#define int __int128
using namespace std;
inline __int128 read()
{
int X=0,w=0; char ch=0;
while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
return w?-X:X;
}
void print(__int128 x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)print(x/10);
putchar(x%10+'0');
}
signed main() {
signed t;
cin >> t;
while(t--) {
int n;
n = read();
//邊平行於棱的肯定沒有
if(n == 0 || n == 1) {
cout << 0 << endl;
continue;
}
n--;
//cout << (1 + n) * n / 2 * 8 << endl;
n %= mod;
print((1 + n) * n % mod * (1 + n) % mod * n % mod * 2 % mod);
cout << endl;
}
return 0;
}