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

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

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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;
}