Problem 1692 Key problem
Accept: 103 Submit: 553 Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Whenever rxw meets Coral, he requires her to give him the laboratory key. Coral does not want to give him the key, so Coral ask him one question. if rxw can solve the problem, she will give him the key, otherwise do not give him. rxw turns to you for help now,can you help him?N children form a circle, numbered 0,1,2, ... ..., n-1,with Clockwise. Initially the ith child has Ai apples. Each round game, the ith child will obtain ( L*A(i+n-1)%n+R*A(i+1)%n ) apples. After m rounds game, Coral would like to know the number of apples each child has. Because the final figure may be very large, so output the number model M.
Input
The first line of input is an integer T representing the number of test cases to follow. Each case consists of two lines of input: the first line contains five integers n,m,L,R and M . the second line contains n integers A0, A1, A2 ... An-1. (0 <= Ai <= 1000,3 <= n <= 100,0 <= L, R <= 1000,1 <= M <= 10 ^ 6,0 <=m < = 10 ^ 9). After m rounds game, output the number model M of apples each child has.
Output
Each case separated by a space. See sample.
Sample Input
1
3 2 3 4 10000
1 2 3
Sample Output
120 133 131
Source
FOJ月赛-2009年3月--- Coral矩阵构造，题意有问题。R,L正好反了。
这一题的矩阵，第一次快速幂超时了，很无奈。原因在于Multiply()是f(n^3)， 时间复杂度logm*n^3,
优化的地方，就是在这。
由于构造的矩阵很有规律是同构的。“什么是同构”？
如这一题：
123
312
231
这样的话只要求出第一排，那么其他就很好求了。n^2解决。不会超630ms
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 #define N 102 7 typedef __int64 LL; 8 LL n,m,L,R,mod; 9 LL f[102]; 10 11 struct Matrix 12 { 13 LL mat[N][N]; 14 void ini() 15 { 16 memset(mat,0,sizeof(mat)); 17 } 18 void init() 19 { 20 for(LL i=0;i<n;i++) 21 for(LL j=0;j<n;j++) 22 if(i==j) mat[i][j]=1; 23 else mat[i][j]=0; 24 } 25 void make_first() 26 { 27 for(LL i=0;i<n;i++) 28 { 29 LL k1=(i+n-1)%n; 30 mat[i][k1]=R; 31 LL k2=(i+1)%n; 32 mat[i][k2]=L; 33 } 34 } 35 }M_hxl,M_tom; 36 37 38 Matrix Multiply(Matrix &x, Matrix &y) 39 { 40 LL i, j, k; 41 Matrix z; 42 z.ini(); 43 for(k = 0; k < n; k ++) 44 if(x.mat[0][k]) 45 { 46 for(j = 0; j < n; j ++) 47 if(y.mat[k][j]) 48 z.mat[0][j] = (z.mat[0][j] + (LL)x.mat[0][k] * y.mat[k][j]) % mod; 49 } 50 for(i = 1; i < n; i ++) 51 { 52 z.mat[i][0] = z.mat[i - 1][n - 1]; 53 for(j = 1; j < n; j ++) 54 z.mat[i][j] = z.mat[i - 1][j - 1]; 55 } 56 return z; 57 } 58 void cs() 59 { 60 LL i,j; 61 for(i=0;i<n;i++) 62 { 63 printf(" "); 64 for(j=0;j<n;j++) 65 printf("%I64d ",M_hxl.mat[i][j]); 66 } 67 printf(" "); 68 } 69 70 void power_sum2() 71 { 72 M_hxl.init(); 73 M_hxl.make_first(); 74 M_tom.init(); 75 cs(); 76 while(m) 77 { 78 if(m&1) 79 { 80 M_tom=Multiply(M_tom,M_hxl); 81 } 82 m=m>>1; 83 M_hxl=Multiply(M_hxl,M_hxl); 84 } 85 for(LL i=0;i<n;i++) 86 { 87 LL sum=0; 88 for(LL j=0;j<n;j++) 89 { 90 sum=(sum+f[j]*M_tom.mat[i][j])%mod; 91 } 92 if(i==0)printf("%I64d",sum); 93 else printf(" %I64d",sum); 94 } 95 printf(" "); 96 } 97 98 int main() 99 { 100 LL T; 101 while(scanf("%I64d",&T)>0) 102 { 103 while(T--) 104 { 105 scanf("%I64d%I64d%I64d%I64d%I64d",&n,&m,&L,&R,&mod); 106 for(LL i=0;i<n;i++) 107 scanf("%I64d",&f[i]); 108 if(m==0) 109 { 110 for(LL i=0;i<n;i++) 111 { 112 if(i==0)printf("%I64d",f[i]%mod); 113 else printf(" %I64d",f[i]%mod); 114 } 115 printf(" "); 116 continue; 117 } 118 power_sum2(); 119 } 120 } 121 return 0; 122 }