描述
题解
题意很简单,是求 ⌊an‾‾√⌋ % MOD 。
设 fn=⌊an‾‾√⌋ % MOD ,
首先我们找出前几项看看有啥子规律,
f2=31,f3=197,f4=1255,f5=7997,f6=50959 ,
由题可知,
h2=6,h3=35,h4=190,h5=1267,h6=7862 。
大佬们说,这两个函数递推很相似,根据已知 hn 的递推式,我们可以获悉,
fn=4fn−1+17fn−2−12fn−3 ,
这样,很明显就是一个递推式求第 n <script type="math/tex" id="MathJax-Element-618">n</script> 项的问题了,如此这般,我们直接用矩阵快速幂就好了。
代码
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int MAGIC_1 = 31;
const int MAGIC_2 = 197;
const int MAGIC_3 = 1255;
struct matrix
{
ll v[3][3];
matrix()
{
memset(v, 0, sizeof(v));
}
};
ll n;
matrix M, E, ans;
void init()
{
for (int i = 0; i < 3; i++)
{
E.v[i][i] = 1;
}
M.v[0][0] = 4, M.v[0][1] = 17, M.v[0][2] = -12;
M.v[1][0] = 1, M.v[1][1] = 0, M.v[1][2] = 0;
M.v[2][0] = 0, M.v[2][1] = 1, M.v[2][2] = 0;
}
matrix mul(matrix a, matrix b)
{
matrix c;
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
for (int k = 0; k < 3; k++)
{
c.v[i][j] = (c.v[i][j] % MOD + a.v[i][k] * b.v[k][j] % MOD + MOD) % MOD;
}
}
}
return c;
}
matrix pow(matrix p, ll k)
{
matrix tmp = E;
while (k)
{
if (k & 1)
{
tmp = mul(tmp, p);
k--;
}
k >>= 1;
p = mul(p, p);
}
return tmp;
}
int main()
{
init();
int T;
scanf("%d", &T);
while (T--)
{
scanf("%lld", &n);
if (n == 2)
{
printf("%d\n", MAGIC_1);
}
else if (n == 3)
{
printf("%d\n", MAGIC_2);
}
else if (n == 4)
{
printf("%d\n", MAGIC_3);
}
else
{
ans = pow(M, n - 4);
ll res = (ans.v[0][0] * MAGIC_3 % MOD
+ ans.v[0][1] * MAGIC_2 % MOD
+ ans.v[0][2] * MAGIC_1 % MOD + MOD) % MOD;
printf("%lld\n", res);
}
}
return 0;
}