BZOJ 2326: [HNOI2011]数学作业

自己手推一下然后矩阵就出来了。

可以发现对于矩阵：

\begin{pmatrix}
P_n\\
n\\
1
\end{pmatrix}

左乘矩阵（其中\( 10^k \)代表数位权）：

\begin{pmatrix}
10^k & 1 &1 \\
0& 1 & 1\\
0& 0& 1
\end{pmatrix}

即

 根据矩阵乘法的结合率，对于k分开考虑，进行快速幂计算即可。

有一个问题：能在计算时取模吗？不取模正确性是显然的，取模的话就比较dt了，不好说明答案没有改变。。如果有人知道如何证明取模是正确的，请回复sbit，万谢！

还有一个问题：两个大整数相乘中间结果可能爆掉，两个方法：一是把\( 10^k \)传入时先对m取模，二是模乘法，代码如下：

 

LL mul(LL a, LL b) {
    LL ret = 0;
    for(; b; b >>= 1, a = (a << 1) % mod) {
        if(b & 1) {
            ret = (ret + a) % mod;
        }
    }
    return ret;
}

 

然后。。。没了。。。

  1 //{HEADS
  2 #define FILE_IN_OUT
  3 #define debug
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cmath>
  8 #include <ctime>
  9 #include <algorithm>
 10 #include <iostream>
 11 #include <fstream>
 12 #include <vector>
 13 #include <stack>
 14 #include <queue>
 15 #include <deque>
 16 #include <map>
 17 #include <set>
 18 #include <bitset>
 19 #include <complex>
 20 #include <string>
 21 #define REP(i, j) for (int i = 1; i <= j; ++i)
 22 #define REPI(i, j, k) for (int i = j; i <= k; ++i)
 23 #define REPD(i, j) for (int i = j; 0 < i; --i)
 24 #define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i)
 25 #define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i)
 26 #define CLR(s) memset(s, 0, sizeof s)
 27 #define SET(s, v) memset(s, v, sizeof s)
 28 #define mp make_pair
 29 #define pb push_back
 30 #define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ' '; } cout << endl
 31 #define PS(k) STLR(i, k) { cout << k[i] << ' '; } cout << endl
 32 using namespace std;
 33 #ifdef debug
 34 #ifndef ONLINE_JUDGE
 35     const int OUT_PUT_DEBUG_INFO = 1;
 36 #endif
 37 #endif
 38 #ifdef ONLINE_JUDGE
 39     const int OUT_PUT_DEBUG_INFO = 0;
 40 #endif
 41 #define DG if(OUT_PUT_DEBUG_INFO)
 42 void FILE_INIT(string FILE_NAME) {
 43 #ifdef FILE_IN_OUT
 44 #ifndef ONLINE_JUDGE 
 45     freopen((FILE_NAME + ".in").c_str(), "r", stdin);
 46     freopen((FILE_NAME + ".out").c_str(), "w", stdout);
 47  
 48 #endif
 49 #endif
 50 }
 51 typedef long long LL;
 52 typedef double DB;
 53 typedef pair<int, int> i_pair;
 54 const int INF = 0x3f3f3f3f;
 55 //}
 56  
 57 LL n, p[20];
 58 /* { Matrix Multiplication index int型*/
 59  
 60 const int max_size = 4;
 61 LL mod;
 62 struct Matrix {
 63     LL d[max_size][max_size];
 64     int r, c;
 65     inline Matrix() {
 66         memset(d, 0, sizeof d);
 67     }
 68     inline void set_size(int set_r, int set_c) {
 69         r = set_r;
 70         c = set_c;
 71     }
 72     inline void init(int size) {
 73         r = c = size;
 74         for(int i = 1; i <= r; ++i) {
 75             d[i][i] = 1ull;
 76         }
 77     }
 78     inline void print() {
 79         for(int i = 1; i <= r; ++i) {
 80             for(int j = 1; j < c; ++j) {
 81                 printf("%lld ", d[i][j]);
 82             }
 83             printf("%lld\n", d[i][c]);
 84         }
 85     }
 86 };
 87 LL mul(LL a, LL b) {
 88     LL ret = 0;
 89     for(; b; b >>= 1, a = (a << 1) % mod) {
 90         if(b & 1) {
 91             ret = (ret + a) % mod;
 92         }
 93     }
 94     return ret;
 95 }
 96 inline Matrix operator * (const Matrix &lhs, const Matrix &rhs) {
 97     Matrix ret;
 98     ret.set_size(lhs.r, rhs.c);
 99     for(int i = 1; i <= lhs.r; ++i) {
100         for(int j = 1; j <= rhs.c; ++j) {
101             for(int k = 1; k <= lhs.c; ++k) {
102                 ret.d[i][j] = (ret.d[i][j] + mul(lhs.d[i][k], rhs.d[k][j])) % mod;
103             }
104         }
105     }
106     return ret;
107 }
108  
109 inline Matrix fast_pow(Matrix base, LL index) {
110     Matrix ret;
111     ret.init(base.r);
112     for(; index; index >>= 1, base = base * base) {
113         if(index & 1) {
114             ret = ret * base;
115         }
116     }
117     //ret.print();
118     return ret;
119 }
120      
121  
122 /*} */
123  
124 int main() {
125     FILE_INIT("BZOJ2326");
126  
127     cin >> n >> mod;
128     p[0] = 1;
129     for(int i = 1; i <= 19; ++i) {
130         p[i] = (10 * p[i - 1]);
131     }
132     int len = 0;
133     for(LL tmp = n; tmp; tmp /= 10, ++len);
134     Matrix A, B;
135     A.set_size(3, 1);
136     A.d[1][1] = 0;
137     A.d[2][1] = 0;
138     A.d[3][1] = 1;
139     B.set_size(3, 3);
140     for(int i = 1; i < len; ++i) {
141         B.d[1][1] = p[i] % mod; B.d[1][2] = 1; B.d[1][3] = 1;
142         B.d[2][1] =    0; B.d[2][2] = 1; B.d[2][3] = 1;
143         B.d[3][1] =    0; B.d[3][2] = 0; B.d[3][3] = 1;
144         A = fast_pow(B, p[i - 1] * 9) * A;
145     //  A.print();
146     }
147     B.d[1][1] = p[len] % mod; B.d[1][2] = 1; B.d[1][3] = 1;
148     B.d[2][1] =      0; B.d[2][2] = 1; B.d[2][3] = 1;
149     B.d[3][1] =      0; B.d[3][2] = 0; B.d[3][3] = 1;
150     A = fast_pow(B, n - p[len - 1ull] + 1) * A;
151     cout << A.d[1][1] << endl;
152  
153     return 0;
154 }
View Code

 

posted @ 2014-08-02 23:16  sbit  阅读(265)  评论(0)  编辑  收藏  举报