HDU.1757 A Simple Math Problem (矩阵快速幂)

HDU.1757 A Simple Math Problem (矩阵快速幂)

点我挑战题目

题意分析

给出一个递推式：
1.x<=9时，f(x) = x，
2.x>9时，f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10).
现在想让你求解f(k) % m 的值为多少.

当x<=9时，特判输出x % m.
当x>9是，构造如下矩阵乘法：

f(x) 0 0 0 0 0 0 0 0 0 f(x-1) 0 0 0 0 0 0 0 0 0 f(x-2) 0 0 0 0 0 0 0 0 0 f(x-3) 0 0 0 0 0 0 0 0 0 f(x-4) 0 0 0 0 0 0 0 0 0 f(x-5) 0 0 0 0 0 0 0 0 0 f(x-6) 0 0 0 0 0 0 0 0 0 f(x-7) 0 0 0 0 0 0 0 0 0 f(x-8) 0 0 0 0 0 0 0 0 0 f(x-9) 0 0 0 0 0 0 0 0 0

等于

a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0

乘

f(x-1) 0 0 0 0 0 0 0 0 0 f(x-2) 0 0 0 0 0 0 0 0 0 f(x-3) 0 0 0 0 0 0 0 0 0 f(x-4) 0 0 0 0 0 0 0 0 0 f(x-5) 0 0 0 0 0 0 0 0 0 f(x-6) 0 0 0 0 0 0 0 0 0 f(x-7) 0 0 0 0 0 0 0 0 0 f(x-8) 0 0 0 0 0 0 0 0 0 f(x-9) 0 0 0 0 0 0 0 0 0 f(x-10) 0 0 0 0 0 0 0 0 0

如此便可得到最后结果。

代码总览

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <sstream>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#define INF 0x3f3f3f3f#define nmax 200#define MEM(x) memset(x,0,sizeof(x))using namespace std;const int Dmax = 11;int N = 10;int MOD;typedef struct{    int matrix[Dmax][Dmax];    void init()//初始化为单位矩阵    {        memset(matrix,0,sizeof(matrix));        for(int i = 0; i<Dmax;++i) matrix[i][i] = 1;    }}MAT;MAT ADD(MAT a, MAT b){    for(int i = 0; i<N;++i){        for(int j = 0;j<N;++j){            a.matrix[i][j] +=b.matrix[i][j];            a.matrix[i][j] %= MOD;        }    }    return a;}MAT MUL(MAT a, MAT b){    MAT ans;    for(int i = 0; i<N;++i){        for(int j = 0; j<N;++j){            ans.matrix[i][j] = 0;            for(int k = 0; k<N;++k){                ans.matrix[i][j] += ( (a.matrix[i][k] % MOD) * (b.matrix[k][j] % MOD) ) % MOD;            }            ans.matrix[i][j] %= MOD;        }    }    return ans;}MAT POW(MAT a, int t){    MAT ans; ans.init();    while(t){        if(t&1) ans = MUL(ans,a);        t>>=1;        a = MUL(a,a);    }    return ans;}void OUT(MAT a){    for(int i = 0; i<N;++i){        for(int j =  0; j<N;++j){            printf("%5d",a.matrix[i][j]);        }        printf("\n");    }}void IN(MAT & a,MAT & temp){    memset(a.matrix,0,sizeof(a.matrix));    memset(temp.matrix,0,sizeof(temp.matrix));    for(int i = 0; i<N;++i) a.matrix[i][0] = 9-i;    for(int i = 0; i<N;++i) scanf("%d",&temp.matrix[0][i]);    for(int i = 1; i<N;++i) temp.matrix[i][i-1] = 1;}void CAL(MAT a){    printf("%d\n",a.matrix[0][0] % MOD);}int main(){    //freopen("in.txt","r",stdin);    int K;    while(scanf("%d%d",&K,&MOD) != EOF){        if(K<=9){            printf("%d\n",K);            continue;        }        MAT init,temp;        IN(init,temp);        temp = POW(temp,K-9);        temp = MUL(temp,init);        CAL(temp);    }    return 0;}