这是本人写的第一次博客，学了半年的基础C语言，初学算法，若有错误还请指正。
E. Tetrahedron
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a tetrahedron. Let's mark its vertices with letters A, B, C and D correspondingly.
An ant is standing in the vertex D of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex D to itself in exactly n steps. In other words, you are asked to find out the number of different cyclic paths with the length of n from vertex D to itself. As the number can be quite large, you should print it modulo 1000000007 (109 + 7).
Input
The first line contains the only integer n (1 ≤ n ≤ 107) — the required length of the cyclic path.
Output
Print the only integer — the required number of ways modulo 1000000007 (109 + 7).
Sample test(s)
input
2
output
3
input
4
output
21
Note
The required paths in the first sample are:
D - A - D
D - B - D
D - C - D
;则第n次到达D点的方法数等于总数(3 ^ n)乘以概率
，设第n次到达D点的方法数为dp[n], 则有
;这个可以直接用暴力写，也可以用矩阵快速幂写，在这就不贴代码了，不过高中学过数学竞赛的同学可能会觉得这个式子结构()有点眼熟, 3和-1的次数都是等次幂，让人联想到线性递推数列的求解 : 对于数列An = p * An-1 + q *An-2, 其特征方程为x^2 = px + q, 假设其两根分别为x1, x2,则
(x1, x2不相等时)或
(x1 = x2时)，a,b为常数，可由给出条件获得，所以可以把
中的3和-1分别看成x1,x2,由二次方程(设为x^2 + px + q = 0)的韦达定理的p = -(x1 + x2) = -2, q = x1 *x2 = 3;即方程为x^2 - 2x - 3 = 0,而此方程为数列An = 2 * An-1 + 3 *An-2#include <cstdio>
#include <cstring>
const int N = 1e7 + ;
const int mod = 1e9 + ;
long long dp[N], n;
void init(){
dp[] = ;
dp[] = ;
for(int i = ; i < N; i ++){
dp[i] = * dp[i - ] + * dp[i - ];
dp[i] %= mod;
}
}
int main(){
init();
while(scanf("%d", &n) == )
printf("%d\n", dp[n]);
return ;
}
不过刚开始这点思路非常复杂，所有笔者就去想dp[n] = 2 * dp[n - 1] + 3 *dp[n - 2]的含义
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod = 1e9 + ;
int n;
long long a, b;
int main(){
while(scanf("%d", &n) == ){
if(n < )
printf("0\n");
else if(n == )
printf("3\n");
else{
a = ;
b = ;
for(int i = ; i <= n - ; i ++){
if(i & )
a = ( * a + * b)%mod;
else
b = ( * b + * a)%mod;
}
if(n & )
printf("%I64d\n", a);
else
printf("%I64d\n", b);
}
}
return ;
}
#include <cstdio>
#include <cstring>
typedef long long lld;
const int N = 1e7 + ;
const int mod = 1e9 + ;
int dp[N][], n;
void init(){
dp[][] = ;
dp[][] = ;
dp[][] = ;
for(int i = ; i < N; i ++){
for(int j = ; j < ; j ++){
for(int k = ; k < ; k ++){
if(k == j )
continue;
dp[i][j] += dp[i - ][k];
dp[i][j] %= mod;
}
}
}
}
int main(){
init();
while(scanf("%d", &n) == )
printf("%d\n", dp[n][]);
return ;
}
其实可以从这个递推关系式可以推出dp[n] = 2 * dp[n - 1] + 3 *dp[n - 2]
，便可得出另一个关系式:#include <cstdio>
#include <cstring>
const int N = 1e7 + ;
const int mod = 1e9 + ;
long long dp[N], n;
void init(){
dp[] = ;
dp[] = ;
for(int i = ; i < N; i ++){
if(i & )
dp[i] = * dp[i - ] - ;
else
dp[i] = * dp[i - ] + ;
dp[i] %= mod;
}
}
int main(){
init();
while(scanf("%d", &n) == )
printf("%d\n", dp[n]);
return ;
}