HDU 1757 A Simple Math Problem（构造矩阵）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5125    Accepted Submission(s): 3103

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

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场 

 

POINT：

构造两个矩阵

乘

#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;#define  LL long longconst int N = 10;int p;struct mx{    int a[N+1][N+1];};mx cheng(mx a,mx b){    mx ans;    for(int i=1;i<=N;i++)    {        for(int j=1;j<=N;j++)        {            ans.a[i][j]=0;            for(int k=1;k<=N;k++)            {                (ans.a[i][j]+=a.a[i][k]*b.a[k][j])%=p;            }        }    }    return ans;}mx qkm(mx base,int mi){    mx ans;    for(int i=1;i<=N;i++)    {        for(int j=1;j<=N;j++)        {            ans.a[i][j]=i==j;        }    }    while(mi)    {        if(mi&1) ans=cheng(ans,base);        base=cheng(base,base);        mi>>=1;    }    return ans;}int main(){    int k,m;    int a[10];    int f[10];    while(~scanf("%d %d",&k,&m))    {        p=m;        for(int i=0;i<=9;i++)        {            scanf("%d",&a[i]);            f[i]=i;        }        if(k<=9)        {            printf("%d\n",f[k]%m);        }        else        {            mx base;            for(int i=1;i<=10;i++)            {                base.a[1][i]=a[i-1];            }            for(int i=2;i<=10;i++)            {                for(int j=1;j<=10;j++)                {                    if(j==i-1)                    {                        base.a[i][j]=1;                    }                    else base.a[i][j]=0;                }            }            base=qkm(base,k-9);            int ans=0;            for(int k=1;k<=10;k++)            {                (ans+=base.a[1][k]*f[10-k])%=p;            }            printf("%d\n",ans);        }            }}