>Link
ssl 2514
>Description
求序列 f n − f 3 − f 4 − f 5 − . . . − f n − 3 − f n − 2 = ( n + 4 ) ( n − 1 ) / 2 f_n-f_3-f_4-f_5-...-f_{n-3}-f_{n-2}=(n+4)(n-1)/2 fn−f3−f4−f5−...−fn−3−fn−2=(n+4)(n−1)/2 中前 n n n项之和
其中 f 1 = f 2 = 1 f_1=f_2=1 f1=f2=1
>解题思路
看到这个序列公式，我们首先把它给化简一下
f n − f 3 − f 4 − f 5 − . . . − f n − 3 − f n − 2 = ( n + 4 ) ( n − 1 ) / 2 f_n-f_3-f_4-f_5-...-f_{n-3}-f_{n-2}=(n+4)(n-1)/2 fn−f3−f4−f5−...−fn−3−fn−2=(n+4)(n−1)/2
f n = ( n + 4 ) ( n − 1 ) 2 + f 3 + f 4 + f 5 + . . . + f n − 3 + f n − 2 f_n=\frac{(n+4)(n-1)}{2}+f_3+f_4+f_5+...+f_{n-3}+f_{n-2} fn=2(n+4)(n−1)+f3+f4+f5+...+fn−3+fn−2
= n 2 + 3 n − 4 2 + f 3 + f 4 + f 5 + . . . + f n − 3 + f n − 2 =\frac{n^2+3n-4}{2}+f_3+f_4+f_5+...+f_{n-3}+f_{n-2} =2n2+3n−4+f3+f4+f5+...+fn−3+fn−2
从 f n → f n + 1 f_n→f_{n+1} fn→fn+1可推
f n + 1 = ( n + 5 ) n 2 + f 3 + f 4 + f 5 + . . . + f n − 2 + f n − 1 f_{n+1}=\frac{(n+5)n}{2}+f_3+f_4+f_5+...+f_{n-2}+f_{n-1} fn+1=2(n+5)n+f3+f4+f5+...+fn−2+fn−1
= n 2 + 3 n − 4 + 2 n + 4 2 + f 3 + f 4 + f 5 + . . . + f n − 2 + f n − 1 =\frac{n^2+3n-4+2n+4}{2}+f_3+f_4+f_5+...+f_{n-2}+f_{n-1} =2n2+3n−4+2n+4+f3+f4+f5+...+fn−2+fn−1
= ( n 2 + 3 n − 4 2 ) + ( n + 2 ) + ( f 3 + f 4 + f 5 + . . . + f n − 2 ) + ( f n − 1 ) =(\frac{n^2+3n-4}{2})+(n+2)+(f_3+f_4+f_5+...+f_{n-2})+(f_{n-1}) =(2n2+3n−4)+(n+2)+(f3+f4+f5+...+fn−2)+(fn−1)
= f n + f n − 1 + n + 2 =f_{n}+f_{n-1}+n+2 =fn+fn−1+n+2
因此得出 f n = f n − 2 + f n − 1 + n + 1 f_n=f_{n-2}+f_{n-1}+n+1 fn=fn−2+fn−1+n+1
这样这道题就与斐波那契数列Ⅳ相同了
>代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
const LL p = 1000000007;
struct matrix
{
int x, y;
LL a[10][10];
} A, B, ans;
matrix operator *(matrix a, matrix b)
{
matrix c;
c.x = a.x, c.y = b.y;
for (int i = 1; i <= c.x; i++)
for (int j = 1; j <= c.y; j++) c.a[i][j] = 0;
for (int k = 1; k <= a.y; k++)
for (int i = 1; i <= c.x; i++)
for (int j = 1; j <= c.y; j++)
c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j] % p) % p;
return c;
}
void power (LL k)
{
if (k == 1) {
B = A; return;}
power (k >> 1);
B = B * B;
if (k & 1) B = B * A;
}
int main()
{
LL n;
scanf ("%lld", &n);
if (n == 1) {
printf ("1"); return 0;}
A.x = A.y = 5;
A.a[1][2] = A.a[2][1] = A.a[2][2] = A.a[2][3] = A.a[3][3] = A.a[4][2] = 1;
A.a[4][4] = A.a[5][2] = A.a[5][4] = A.a[5][5] = 1;
power (n - 1);
ans.x = 1, ans.y = 5;
ans.a[1][1] = ans.a[1][2] = ans.a[1][3] = ans.a[1][5] = 1;
ans.a[1][4] = 3;
ans = ans * B;
printf ("%lld", ans.a[1][3]);
return 0;
}