Copyright: private individuals do question summed up ~ https://blog.csdn.net/tb_youth/article/details/90941875
// fill the question ~~~
link: https://ac.nowcoder.com/acm/problem/15666
Source: Cattle-off network
Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 32768K, other languages 65536k
64bit the IO the Format: LLD%
Example 1
Input
Copy
4
1
2
3
100
Output
Copy
1
16
57
558616258
/ *
1 matrix structure ~
2 ~ matrix Fast Power
[f(n-2),f(n-1),n^3,n^2,n,1] * A = [f(n-1),f(n),(n+1)^3 ,(n+1)^2,n+1,1]
A = [
0,1,0,0,0,0,
1,1,0,0,0,0,
0,1,1,0,0,0,
0,1,3,1,0,0,
0,1,3,2,1,0,
0,1,1,1,1,1
]
[f(0),f(1),8,4,2,1] * A = [f(1),f(2),27,9,3,1]
[f(0),f(1),8,4,2,1] * A^n = [f(n),f(n=1),f(n+1)^3,f(n+1)^2,n+1,1]
*/
ac_code:
#include <stdio.h>
#define ll long long
const ll mod = 1e9+7;
struct mat
{
ll m[10][10];
}a,e;
mat operator*(const mat x,const mat y)
{
mat ans;
ll tmp;
for(int i = 0; i < 6; i++)
{
for(int j = 0; j < 6; j++)
{
tmp = 0;
for(int k = 0; k < 6; k++)
{
tmp = (tmp%mod+(x.m[i][k]%mod*y.m[k][j]%mod)%mod)%mod;
}
ans.m[i][j] = tmp;
}
}
return ans;
}
mat quickPow(mat a,ll b)
{
mat res = e;
while(b)
{
if(b&1)
res = res*a;
a = a*a;
b >>= 1;
}
return res;
}
int main()
{
for(int i = 0; i < 6; i++)
{
e.m[i][i] = 1;
a.m[i][1] = 1;
a.m[5][i] = 1;
}
a.m[5][0] = 0;
a.m[1][0] = 1;
a.m[2][2] = 1;
a.m[3][2] = 3;
a.m[4][2] = 3;
a.m[3][3] = 1;
a.m[4][3] = 2;
a.m[4][4] = 1;
ll c[10] = {0,1,8,4,2,1};
ll t;
scanf("%lld",&t);
while(t--)
{
ll n;
scanf("%lld",&n);
mat tp = quickPow(a,n);
ll val = 0; //f(n)
for(int i = 0; i < 6; i++)
{
val = (val%mod + tp.m[i][0]*c[i]%mod)%mod;
}
printf("%lld\n",val);
}
return 0;
}