Something about the matrix
HOJ 2255 Not Fibonacci
题目大意:给出一个类似于Fibonacci的递推关系式f(n) = p * f(n-1) + q * f(n - 2).求出
我的做法:根据S(n)和f(n)的递推来构造一个5阶的矩阵:

这样求方阵的n次幂的第一行与初始列向量的乘积再加和即可得到结果。
总结:锻炼递推啊。。
Code:
1
#include <iostream>
2
#include <map>
3
#include <vector>
4
#define N 6
5
using namespace std;
6
struct matrix
7

{
8
long long m[N][N];
9
}E,S;
10
matrix multiply(matrix x, matrix y, long long mod)
11

{
12
matrix tmp;
13
int i, j, k;
14
for(i = 0; i < 5; i++)
15
for(j = 0; j < 5; j++)
16
{
17
tmp.m[i][j] = 0;
18
for(k = 0; k < 5; k++)
19
{
20
tmp.m[i][j] = ((tmp.m[i][j] + x.m[i][k] * y.m[k][j] % mod) + mod) % mod;
21
}
22
}
23
return tmp;
24
}
25
matrix mamod(matrix P, long long b, long long mod)
26

{
27
matrix t = E,y = P;
28
while (b)
29
{
30
if (b & 1)
31
t = multiply(t, y, mod);
32
y = multiply(y, y, mod);
33
b>>=1;
34
}
35
return t;
36
}
37
long long a, b, p, q, s, e;
38
long long Mod = 10000000;
39
40
long long cal(long long n)
41

{
42
if(n == 1)
43
return a + b;
44
if(n == 0)
45
return a;
46
if(n < 0)
47
return 0;
48
matrix ret = mamod(S, n - 1, Mod);
49
50
long long res = 0;
51
long long t[5] =
{a + b, a, b, a, 1};
52
for(int i(0); i < 5; i++)
53
res = ((res + t[i] * ret.m[0][i] % Mod) % Mod + Mod) % Mod;
54
return res;
55
}
56
int main()
57

{
58
int test;
59
scanf("%d", &test);
60
for(int i(0); i < 5; i++)
61
for(int j(0); j < 5; j++)
62
E.m[i][j] = i == j ? 1 : 0;
63
64
while(test--)
65
{
66
scanf("%lld %lld %lld %lld %lld %lld", &a, &b, &p, &q, &s, &e);
67
68
memset(S.m, 0, sizeof(S.m));
69
S.m[1][0] = S.m[3][2] = S.m[4][4] = 1;
70
S.m[0][1] = p + q;
71
S.m[0][2] = p;
72
S.m[2][2] = p;
73
S.m[2][3] = q;
74
S.m[0][4] = a * (1 - p) + b;
75
printf("%lld\n", ((cal(e) - cal(s - 1)) % Mod + Mod ) % Mod);
76
}
77
return 0;
78
}
HOJ 2471 Learning English
题目大意:给出一些长度为2的单词集合S,问可以构造出多少个不同的单词,而且这些单词中的每两个相邻字母构成的2长度单词必须在S中。并给定所求单词的长度上限。
我的做法:问题可以转化为求长为L的通道的条数,每个字母为一个顶点,而且要将这些长度的数目加起来,可以构造矩阵,A为邻接阵,即求S = A + A2 + A3 + … + Ak 然后再 求S中所有数字的和即可。
有两种方法:
1. 用分治的方法来进行二分,S = (E +A + A2 + A3 + … + Ak/2) *(E + Ak/2 )
2. 构造矩阵:

E为单位阵,O为0阵,这样求这个矩阵的n次幂右上角的矩阵则为所求的和矩阵。用快速幂可求.。
Code:
1
#include <iostream>
2
#include <map>
3
#include <vector>
4
#define N 55
5
using namespace std;
6
struct matrix
7

{
8
long long m[N][N];
9
}E,S;
10
matrix multiply(matrix x, matrix y, long long mod)
11

{
12
matrix tmp;
13
int i, j, k;
14
for(i = 0; i < 52; i++)
15
for(j = 0; j < 52; j++)
16
{
17
tmp.m[i][j] = 0;
18
for(k = 0; k < 52; k++)
19
{
20
tmp.m[i][j] = ((tmp.m[i][j] + x.m[i][k] * y.m[k][j] % mod) + mod) % mod;
21
}
22
}
23
return tmp;
24
}
25
matrix mamod(matrix P, long long b, long long mod)
26

{
27
matrix t = E,y = P;
28
while (b)
29
{
30
if (b & 1)
31
t = multiply(t, y, mod);
32
y = multiply(y, y, mod);
33
b>>=1;
34
}
35
return t;
36
}
37
long long s, e;
38
long long Mod = 10000000;
39
char as[213];
40
int main()
41

{
42
int test;
43
scanf("%d", &test);
44
for(int i(0); i < 52; i++)
45
for(int j(0); j < 52; j++)
46
E.m[i][j] = i == j ? 1 : 0;
47
48
while(test--)
49
{
50
scanf("%lld", &s);
51
memset(S.m, 0, sizeof(S.m));
52
for (int i = 0; i < 26; i++)
53
{
54
S.m[i + 26][i + 26] = 1;
55
S.m[i][i + 26] = 1;
56
}
57
for(int i(0); i < s; i++)
58
{
59
scanf("%s", as);
60
S.m[as[0] - 'a'][as[1] - 'a'] = 1;
61
}
62
scanf("%lld", &e);
63
matrix ret = mamod(S, e, Mod);
64
65
long long sum = 0;
66
for (int i = 0; i < 26; i++)
67
{
68
for (int j = 26; j < 52; j++)
69
{
70
sum += ret.m[i][j];
71
if (sum > 1000000)
72
sum %= 1000000;
73
}
74
}
75
printf("%lld\n", sum - 26);
76
}
77
return 0;
78
}
79