HDU：1757 A Simple Math Problem

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4788    Accepted Submission(s): 2876

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

解题思路：

矩阵快速幂题目。

线性代数没白学，好有用，不过这个题目推出矩阵是难点。

别人的图：画的真好，一目了然。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int maxn = 11;int mod;   ///模struct Matrix{    int m[maxn][maxn];}ans,res;///矩阵乘法，计算矩阵a*b。Matrix Mul(Matrix a,Matrix b){   Matrix tmp;   memset(tmp.m,0,sizeof(tmp.m));   for(int i = 1; i <= 10; i++)       for(int j = 1; j <= 10; j++)           for(int k = 1; k <= 10; k++)               tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%mod;   return tmp;}void QuickPower(int N){    ///ans初始为单位矩阵，任何矩阵乘上单位阵都是其本身    memset(ans.m,0,sizeof(ans.m));    for(int i = 1; i <= 10; i++)        ans.m[i][i] = 1;    while(N)    {        if(N&1)            ans = Mul(ans,res);        res = Mul(res,res);        N >>= 1;    }}int main(){    int k;    while(~scanf("%d%d",&k,&mod))    {        memset(res.m,0,sizeof(res.m));        ///输入a[0]到a[9]        for(int i = 1; i <= 10; i++)        {            scanf("%d",&res.m[1][i]);            res.m[1][i] = res.m[1][i]%mod;        }        for(int i = 2; i <= 10; i++)            res.m[i][i-1] = 1;        if(k < 10)            printf("%d\n",k);        else        {            k = k-9;            QuickPower(k);  ///矩阵快速幂            int a = 0;            for(int i = 1; i <= 10; i++)            {                a += ans.m[1][i]*(10-i);                a = a%mod;            }            printf("%d\n",a);        }    }    return 0;}