| 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 |
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: ",s); 129 for(i=0; i<cnt; i++) 130 { 131 if(ans[i]<2) 132 { 133 printf("%lld %lld ",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 ",ak1,ak2); 146 } 147 } 148 }return 0; 149 }