Queuing |
| Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 36 Accepted Submission(s): 29 |
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Problem Description
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 2L 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. |
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Input
Input a length L (0 <= L <= 10 6) and M.
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Output
Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
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Sample Input
3 8 4 7 4 8 |
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Sample Output
6 2 1 |
思路：矩阵乘法，转移就行了
用 f->1 m->0
an 表示 mm
bn 表示 mf
cn 表示 fm
dn 表示 ff
那么很明显 an=cn-1+an-1
bn=an-1 因为不能由 cn-1 fm->fmf
cn=bn-1+dn
dn=bn;
构造出矩阵即可！
1 /* 2 Author:wuhuajun 3 */ 4 #include <cmath> 5 #include <cstdio> 6 #include <algorithm> 7 #include <cstring> 8 #include <string> 9 #include <cstdlib> 10 using namespace std; 11 12 typedef long long ll; 13 typedef double dd; 14 const int maxn=210; 15 16 struct matrix 17 { 18 ll m[5][5]; 19 } c,base,ans; 20 ll ans1,mod,r,t; 21 int n,l; 22 23 void close() 24 { 25 exit(0); 26 } 27 28 matrix mul(matrix a,matrix b) 29 { 30 memset(c.m,0,sizeof(c.m)); 31 for (int i=1;i<=4;i++) 32 for (int j=1;j<=4;j++) 33 for (int k=1;k<=4;k++) 34 { 35 c.m[i][j]+=a.m[i][k]*b.m[k][j]; 36 c.m[i][j] %= mod; 37 } 38 return c; 39 } 40 /* 41 void print(matrix a) 42 { 43 for (int i=1;i<=4;i++) 44 { 45 for (int j=1;j<=4;j++) 46 printf("%lld ",a.m[i][j]); 47 puts(""); 48 } 49 } 50 */ 51 void ass(int x,int y,matrix &base) 52 { 53 base.m[x][y]=1; 54 } 55 56 void work() 57 { 58 memset(base.m,0,sizeof(base.m)); 59 ass(1,2,base);ass(1,3,base); 60 ass(2,4,base); 61 ass(3,2,base); 62 ass(4,1,base);ass(4,4,base); 63 memset(ans.m,0,sizeof(ans.m)); 64 for (int i=1;i<=4;i++) 65 ans.m[i][i]=1; 66 n=l; n-=2; 67 while (n!=0) 68 { 69 if (n & 1) 70 ans=mul(base,ans); 71 n/=2; 72 base=mul(base,base); 73 } 74 for (int i=1;i<=4;i++) 75 for (int j=1;j<=4;j++) 76 ans1+=ans.m[i][j]; 77 /* 78 t=0; 79 for (int i=1;i<=4;i++) 80 t+=ans.m[6][i]; 81 t %= mod; 82 ans1-=t; 83 for (int i=1;i<=5;i++) 84 ans1+=ans.m[5][i]; 85 */ 86 } 87 88 void init() 89 { 90 while (scanf("%d %I64d",&l,&mod)!=EOF) 91 { 92 ans1=0; 93 if (n==1 && n==2) 94 ans1=0; 95 /* 96 else 97 { 98 while (n!=0) 99 { 100 if (n & 1) 101 ans1=(ans1 * r) % mod; 102 n/=2; 103 r=(r * r) % mod; 104 } 105 work(); 106 } 107 */ 108 work(); 109 ans1+=mod; 110 ans1 %= mod; 111 printf("%I64d ",ans1); 112 } 113 } 114 115 int main () 116 { 117 init(); 118 close(); 119 return 0; 120 }