【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。

  1 /* 3509 */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23
 24 // #define DEBUG
 25 #define sti                set<int>
 26 #define stpii            set<pair<int, int> >
 27 #define mpii            map<int,int>
 28 #define vi                vector<int>
 29 #define pii                pair<int,int>
 30 #define vpii            vector<pair<int,int> >
 31 #define rep(i, a, n)     for (int i=a;i<n;++i)
 32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 33 #define clr                clear
 34 #define pb                 push_back
 35 #define mp                 make_pair
 36 #define fir                first
 37 #define sec                second
 38 #define all(x)             (x).begin(),(x).end()
 39 #define SZ(x)             ((int)(x).size())
 40 #define lson            l, mid, rt<<1
 41 #define rson            mid+1, r, rt<<1|1
 42
 43 typedef struct mat_t {
 44     __int64 m[55][55];
 45
 46     mat_t() {
 47         memset(m, 0, sizeof(m));
 48     }
 49 } mat_t;
 50
 51 const int maxn = 55;
 52 __int64 A[maxn], B[maxn], F1[maxn], F2[maxn];
 53 int mod, L;
 54
 55 mat_t matMult(mat_t a, mat_t b) {
 56     mat_t c;
 57
 58     rep(k, 0, L) {
 59         rep(i, 0, L) {
 60             if (a.m[i][k]) {
 61                 rep(j, 0, L) {
 62                     if (b.m[k][j]) {
 63                         c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]%mod)%mod;
 64                     }
 65                 }
 66             }
 67         }
 68     }
 69
 70     return c;
 71 }
 72
 73 mat_t matPow(mat_t a, int n) {
 74     mat_t ret;
 75
 76     rep(i, 0, L)    ret.m[i][i] = 1;
 77
 78     while (n) {
 79         if (n & 1)
 80             ret = matMult(ret, a);
 81         a = matMult(a, a);
 82         n >>= 1;
 83     }
 84
 85     return ret;
 86 }
 87
 88 int main() {
 89     ios::sync_with_stdio(false);
 90     #ifndef ONLINE_JUDGE
 91         freopen("data.in", "r", stdin);
 92         freopen("data.out", "w", stdout);
 93     #endif
 94
 95     int t;
 96     int f1, f2, a, b;
 97     int n, k_;
 98     __int64 ans;
 99     mat_t e, tmp;
100     int i, j, k;
101     __int64 c;
102     int l, r, nc;
103
104     scanf("%d", &t);
105     while (t--) {
106         scanf("%d %d %d %d %d %d %d", &f1,&f2, &a,&b, &k_, &n, &mod);
107         L = k_ + 2;
108         A[0] = B[0] = F1[0] = F2[0] = 1;
109         for (i=1; i<L; ++i) {
110             A[i] = A[i-1] * a % mod;
111             B[i] = B[i-1] * b % mod;
112             F1[i] = F1[i-1] * f1 % mod;
113             F2[i] = F2[i-1] * f2 % mod;
114         }
115
116         memset(e.m, 0, sizeof(e.m));
117         e.m[0][0] = e.m[1][0] = 1;
118         for (j=1,k=k_; j<L; ++j,--k) {
119             for (i=1,c=1,nc=k+1,r=k,l=1; nc; ++i,--nc,c=c*r/l,--r,++l) {
120                 e.m[i][j] = (c % mod) * A[k+1-i] % mod * B[i-1] % mod;
121             }
122         }
123         #ifdef DEBUG
124         for (i=0; i<L; ++i) {
125             for (j=0; j<L; ++j)
126                 printf("%I64d ", e.m[i][j]);
127             putchar(‘\n‘);
128         }
129         #endif
130
131         tmp = matPow(e, n-1);
132
133         ans = F1[k_] * tmp.m[0][0] % mod;
134         for (j=1; j<L; ++j) {
135             ans = (ans + F2[k_+1-j]*F1[j-1]%mod*tmp.m[j][0]%mod)%mod;
136         }
137
138         printf("%I64d\n", ans);
139     }
140
141     #ifndef ONLINE_JUDGE
142         printf("time = %d.\n", (int)clock());
143     #endif
144
145     return 0;
146 }

【HDOJ】3509 Buge's Fibonacci Number Problem