1131 - Just Two Functions

1131 - Just Two Functions
Time Limit: 2 second(s) Memory Limit: 32 MB

Let

fn = a1 * fn-1 + b1 * fn-2 + c1 * gn-3

gn = a2 * gn-1 + b2 * gn-2 + c2 * fn-3

Find fn % M and gn % M. (% stands for the modulo operation.)

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a blank line. Next line contains three integers a1 b1 c1 (0 ≤ a1, b1, c1 < 25000). Next line contains three integers a2 b2 c2 (0 ≤ a2, b2, c2 < 25000). Next line contains three integers f0 f1 f2(0 ≤ f0, f1, f2 < 25000). Next line contains three integers g0 g1 g2 (0 ≤ g0, g1, g2 < 25000). The next line contains an integer M (1 ≤ M < 25000).

Next line contains an integer q (1 ≤ q ≤ 100) denoting the number of queries. Next line contains q space separated integers denoting n. Each of these integers is non-negative and less than 231.

Output

For each case, print the case number in a line. Then for each query, you have to print one line containing fn % M and gn % M.

Sample Input

Output for Sample Input

2

 

1 1 0

0 0 0

0 1 1

0 0 0

20000

10

1 2 3 4 5 6 7 8 9 10

 

1 1 1

1 1 1

2 2 2

2 2 2

20000

5

2 4 6 8 10

Case 1:

1 0

1 0

2 0

3 0

5 0

8 0

13 0

21 0

34 0

55 0

Case 2:

2 2

10 10

34 34

114 114

386 386

 

PROBLEM SETTER: JANE ALAM JAN
思路：矩阵快速幂；
比较简单，矩阵也比较好推。
  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<math.h>
  8 #include<vector>
  9 using namespace std;
 10 typedef long long LL;
 11 char str[100];
 12 char ask[100];
 13 LL ans[200];
 14 int M;
 15 typedef struct pp
 16 {
 17     LL m[10][10];
 18     pp()
 19     {
 20         memset(m,0,sizeof(m));
 21     }
 22 } maxtr;
 23 LL xishu[10];
 24 LL xisu[10];
 25 LL f[10];
 26 LL g[10];
 27 maxtr E()
 28 {
 29     maxtr ac;
 30     int i,j;
 31     for(i=0; i<10; i++)
 32     {
 33         for(j=0; j<10; j++)
 34         {
 35             if(i==j)
 36             {
 37                 ac.m[i][j]=1;
 38             }
 39             else ac.m[i][j]=0;
 40         }
 41     }
 42     return ac;
 43 }
 44 void Init(maxtr *p)
 45 {
 46     int i,j,k;
 47     memset(p->m,0,sizeof(p->m));
 48     p->m[0][0]=xishu[0];
 49     p->m[0][1]=xishu[1];
 50     p->m[0][5]=xishu[2];
 51     p->m[1][0]=1;
 52     p->m[2][1]=1;
 53     p->m[3][2]=xisu[2];
 54     p->m[3][3]=xisu[0];
 55     p->m[3][4]=xisu[1];
 56     p->m[4][3]=1;
 57     p->m[5][4]=1;
 58 }
 59 maxtr quick(maxtr C,LL m)
 60 {
 61     maxtr ak=E();
 62     int s;
 63     int i,j;
 64     while(m)
 65     {
 66         if(m&1)
 67         {
 68             maxtr vv;
 69             memset(vv.m,0,sizeof(vv.m));
 70             for(i=0; i<=5; i++)
 71             {
 72                 for(j=0; j<=5; j++)
 73                 {
 74                     for(s=0; s<=5; s++)
 75                     {
 76                         vv.m[i][j]=(vv.m[i][j]+C.m[i][s]*ak.m[s][j]%M)%M;
 77                     }
 78                 }
 79             }
 80             ak=vv;
 81         }
 82         maxtr vv;memset(vv.m,0,sizeof(vv.m));
 83         for(i=0; i<=5; i++)
 84         {
 85             for(j=0; j<=5; j++)
 86             {
 87                 for(s=0; s<=5; s++)
 88                 {
 89                     vv.m[i][j]=(vv.m[i][j]+C.m[i][s]*C.m[s][j]%M)%M;
 90                 }
 91             }
 92         }
 93         C=vv;
 94         m/=2;
 95     }
 96     return ak;
 97 }
 98 int main(void)
 99 {
100     LL i,j,k;
101     scanf("%lld",&k);
102     LL s;
103     for(s=1; s<=k; s++)
104     {
105         for(i=0; i<3; i++)
106         {
107             scanf("%lld",&xishu[i]);
108         }
109         for(i=0; i<3; i++)
110         {
111             scanf("%lld",&xisu[i]);
112         }
113         for(i=0; i<3; i++)
114         {
115             scanf("%lld",&f[i]);
116         }
117         for(i=0; i<3; i++)
118         {
119             scanf("%lld",&g[i]);
120         }
121         scanf("%d",&M);
122         int cnt=0;
123         scanf("%d",&cnt);
124         for(i=0; i<cnt; i++)
125         {
126             scanf("%lld",&ans[i]);
127         }
128         printf("Case %d:\n",s);
129         for(i=0; i<cnt; i++)
130         {
131             if(ans[i]<2)
132             {
133                 printf("%lld %lld\n",f[ans[i]]%M,g[ans[i]]%M);
134             }
135             else
136             {
137                 maxtr ac;
138                 memset(ac.m,0,sizeof(ac.m));
139                 Init(&ac);
140                 maxtr ak=quick(ac,ans[i]-2);
141                 LL ak1=ak.m[0][0]*f[2]%M+ak.m[0][1]*f[1]%M+ak.m[0][5]*g[0]%M+ak.m[0][2]*f[0]%M+ak.m[0][3]*g[2]%M+ak.m[0][4]*g[1]%M;
142                 ak1%=M;
143                 LL ak2=ak.m[3][2]*f[0]%M+ak.m[3][3]*g[2]%M+ak.m[3][4]*g[1]%M+ak.m[3][0]*f[2]%M+ak.m[3][1]*f[1]%M+ak.m[3][5]*g[0]%M;
144                 ak2%=M;
145                 printf("%lld %lld\n",ak1,ak2);
146             }
147         }
148     }return 0;
149 }

 

posted @ 2016-06-17 10:16  sCjTyC  阅读(232)  评论(0)  编辑  收藏  举报