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 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  

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;}