Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.
Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2 L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.
Input
Input a length L (0 <= L <= 10 6) and M.
Output
Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
Sample Input
3 8
4 7
4 8
Sample Output
6
2
1
Your task is to calculate the number of E-queues mod M with length L by writing a program.
Input
Input a length L (0 <= L <= 10 6) and M.
Output
Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
Sample Input
3 8
4 7
4 8
Sample Output
6
2
1
题意：就是给你这个字符串的长度L，这个字符串只能由f和m组成，让你用这两个字母来构造各种字符串，但是里面不能出现fmf或fff，求这个L长度里面不出现fmf和fff的字符串个数，最后在取余m
我感觉这个题拿到的时候不知道怎么去下手，不像之前的快速幂，那些递推公式题目相当于直接告诉你了，但是这个还要推出来
既然是快速幂做的，那么肯定是后项是由前项有某种关系组成的
例如已知长度为7的串的答案，那么8怎么推出来
8就比7多出来一个字母，我们那么8可以看作在7后面加了一个字符
当在7后面加一个m的话，那么就不会影响原串
但是如果加一个f，那么8这个串后面三个之能是mmf、mff
当为mmf的话，相当于这三个不会受到前面字符的影响，相应的它也不会影响前面的字符
当为mff倒数第四个只能是m，二且这样的话也不会影响前面
所以这样递推公式就出来了
| 1 0 1 1 | * | F(n-1) | = | F(n) |
| 1 0 0 0 | | F(n-2) | | F(n-1) |
| 0 1 0 0 | | F(n-3) | | F(n-2) |
| 0 0 1 0 | | F(n-4) | | F(n-3) |
上代码：
1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 const int maxn=15; 7 typedef long long ll; 8 ll n=7,mod; 9 ll e[maxn][maxn],v[maxn][maxn],w[maxn][maxn]; 10 void mul(ll a[maxn][maxn],ll b[maxn][maxn]) 11 { 12 13 memset(e,0,sizeof(e)); 14 for(ll i=1;i<=n;++i) 15 { 16 for(ll j=1;j<=n;++j) 17 { 18 ll sum=0; 19 for(ll k=1;k<=n;++k) 20 { 21 sum=(sum%mod+(a[i][k]*b[k][j])%mod)%mod; 22 } 23 e[i][j]=sum; 24 } 25 } 26 for(ll i=1;i<=n;++i) 27 { 28 for(ll j=1;j<=n;++j) 29 { 30 a[i][j]=e[i][j]; 31 } 32 } 33 } 34 void pow(ll b) 35 { 36 for(ll i=1;i<=n;++i) 37 { 38 w[i][i]=1; 39 } 40 while(b) 41 { 42 if(b&1) 43 { 44 mul(w,v); 45 } 46 mul(v,v); 47 b>>=1; 48 } 49 } 50 int main() 51 { 52 ll a,b; 53 while(~scanf("%lld%lld",&a,&b)) 54 { 55 mod = b; 56 memset(w,0,sizeof(w)); 57 memset(v,0,sizeof(v)); 58 if(a==1) 59 { 60 int c=2%mod; 61 printf("%d ",c); 62 continue; 63 } 64 else if(a==2) 65 { 66 int c=4%mod; 67 printf("%d ",c); 68 continue; 69 } 70 else if(a==3) 71 { 72 int c=6%mod; 73 printf("%d ",c); 74 continue; 75 } 76 else if(a==4) 77 { 78 int c=9%mod; 79 printf("%d ",c); 80 continue; 81 } 82 else if(a==0) 83 { 84 printf("0 "); 85 continue; 86 } 87 //0 2 4 6 9 88 v[1][1]=1;v[1][2]=0;v[1][3]=1;v[1][4]=1; 89 v[2][1]=1;v[2][2]=0;v[2][3]=0;v[2][4]=0; 90 v[3][1]=0;v[3][2]=1;v[3][3]=0;v[3][4]=0; 91 v[4][1]=0;v[4][2]=0;v[4][3]=1;v[4][4]=0; 92 pow(a-4); 93 ll sum=(w[1][1]*9+w[1][2]*6+w[1][3]*4+w[1][4]*2)%mod; 94 printf("%lld ",sum); 95 } 96 return 0; 97 }