原题http://acm.hdu.edu.cn/showproblem.php?pid=1757

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2526    Accepted Submission(s): 1469

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

45
104

//本题只要把矩阵构造出来，接下来的就是套模板了
//构建的矩阵相乘的类型为AAAAAAB，B为已知的函数前几项
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
using namespace std;
#define max 15
int k,mod;

struct Matrix{
	int m[max][max];
};

Matrix unit,init;

void Init(){
	int i;
	memset(init.m,0,sizeof(init.m));
	for(i=1;i<10;i++){
		init.m[i][i-1] = 1;//自己构建的那个矩阵
	}
	memset(unit.m,0,sizeof(unit.m));
	for(i=0;i<10;i++){
		unit.m[i][i] = 1;
	}
}

Matrix Mul(Matrix a,Matrix b){
	Matrix c;
	int i,j,k;
	for(i=0;i<10;i++){
		for(j=0;j<10;j++){
			c.m[i][j] = 0;
			for(k=0;k<10;k++){
				c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
			}
			c.m[i][j]%=mod;
		}
	}
	
	return c;
}

Matrix Pow(Matrix a,Matrix b,int x){
	while(x){
		if(x&1){
			b = Mul(a,b);
		}
		a = Mul(a,a);
		x>>=1;
	}
	
	return b;
}

int main(){
	int i;

	while(~scanf("%d%d",&k,&mod)){
		Init();
		for(i=0;i<10;i++){
			scanf("%d",&init.m[0][i]);
		}
		if(k < 10){
			printf("%d\n",k%mod);
		}
		Matrix res = Pow(init,unit,k-9);
		int ans = 0;
		for(i=0;i<10;i++){
			ans+=(res.m[0][i]*(9-i))%mod;//因为这个函数的前10项为9,8,7,6,5,4,3,2,1,0
		}
		ans%=mod;
		printf("%d\n",ans);
	}

	return 0;
}

HDU1757矩阵的简单运用,布布扣,bubuko.com

HDU1757矩阵的简单运用

原文：http://blog.csdn.net/zcr_7/article/details/38257775