HDU1757矩阵的简单运用

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

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

 

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

 

Output

For each case, output f(k) % m in one line.

 

Sample Input

10 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

 

Sample Output

45104

//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 15int 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;}