hdoj.1757 A Simple Math Problem【矩阵快速幂】 2015/07/28

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3457    Accepted Submission(s): 2085

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

 

Sample Output

  
  

   
   45
104
  
  

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场  

注：矩阵快速幂模板，关键是矩阵不好建。。。

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

struct node{
    int map[10][10];
}per,a;

int k,m;

node cheng(node x,node y){
    node c;
    for( int i = 0 ; i < 10 ; ++i ){
        for( int j = 0 ; j < 10 ; ++j ){
            c.map[i][j] = 0;
            for( int k = 0 ; k < 10 ; ++k )
                c.map[i][j] += ( x.map[i][k] * y.map[k][j] ) % m;
            c.map[i][j] %= m;
        }
    }
    return c;
}

void modefy(){
    node ans = per;
    while( k ){
        if( k & 1 )
            ans = cheng(ans,a);
        a = cheng(a,a);
        k >>= 1;
    }
    int sum = 0;
    for( int i = 0 ; i < 10 ; ++i )
        sum += ans.map[0][i] * (9-i);
    printf("%d\n",sum%m);
}

int main(){
    while( ~scanf("%d %d",&k,&m) ){
        if( k < 10 ){
            printf("%d\n",k);
            continue;
        }
        memset( per.map,0,sizeof(per.map) );
        memset( a.map,0,sizeof(a.map) );
        for( int i = 1 ; i < 10 ; ++i )
            a.map[i][i-1] = 1;
        for( int i = 0 ; i < 10 ; ++i ){
            scanf("%d",&a.map[0][i]);
            per.map[i][i] = 1;
        }
        k -= 9;
        modefy();
    }
    return 0;
}