【HDOJ】3509 Buge's Fibonacci Number Problem

快速矩阵幂，系数矩阵由多个二项分布组成。
第1列是（0，（a+b）^k）
第2列是（0，（a+b）^(k-1),0）
第3列是（0，（a+b）^(k-2),0，0）
以此类推。

 /* 3509 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 // #define DEBUG
 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct mat_t {
     __int64 m[][];

     mat_t() {
         memset(m, , sizeof(m));
     }
 } mat_t;

 const int maxn = ;
 __int64 A[maxn], B[maxn], F1[maxn], F2[maxn];
 int mod, L;

 mat_t matMult(mat_t a, mat_t b) {
     mat_t c;

     rep(k, , L) {
         rep(i, , L) {
             if (a.m[i][k]) {
                 rep(j, , L) {
                     if (b.m[k][j]) {
                         c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]%mod)%mod;
                     }
                 }
             }
         }
     }

     return c;
 }

 mat_t matPow(mat_t a, int n) {
     mat_t ret;

     rep(i, , L)    ret.m[i][i] = ;

     while (n) {
         if (n & )
             ret = matMult(ret, a);
         a = matMult(a, a);
         n >>= ;
     }

     return ret;
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int t;
     int f1, f2, a, b;
     int n, k_;
     __int64 ans;
     mat_t e, tmp;
     int i, j, k;
     __int64 c;
     int l, r, nc;

     scanf("%d", &t);
     while (t--) {
         scanf("%d %d %d %d %d %d %d", &f1,&f2, &a,&b, &k_, &n, &mod);
         L = k_ + ;
         A[] = B[] = F1[] = F2[] = ;
         for (i=; i<L; ++i) {
             A[i] = A[i-] * a % mod;
             B[i] = B[i-] * b % mod;
             F1[i] = F1[i-] * f1 % mod;
             F2[i] = F2[i-] * f2 % mod;
         }

         memset(e.m, , sizeof(e.m));
         e.m[][] = e.m[][] = ;
         for (j=,k=k_; j<L; ++j,--k) {
             for (i=,c=,nc=k+,r=k,l=; nc; ++i,--nc,c=c*r/l,--r,++l) {
                 e.m[i][j] = (c % mod) * A[k+-i] % mod * B[i-] % mod;
             }
         }
         #ifdef DEBUG
         for (i=; i<L; ++i) {
             for (j=; j<L; ++j)
                 printf("%I64d ", e.m[i][j]);
             putchar('\n');
         }
         #endif

         tmp = matPow(e, n-);

         ans = F1[k_] * tmp.m[][] % mod;
         for (j=; j<L; ++j) {
             ans = (ans + F2[k_+-j]*F1[j-]%mod*tmp.m[j][]%mod)%mod;
         }

         printf("%I64d\n", ans);
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }