A Simple Math Problem （谢庆皇）

/*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
*/
/*                             A Simple Math Problem
Time limit: 1000MS Memory limit: 32768K
Total Submit: 5 Accepted: 4

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*109 , m < 104 )
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

解题的关键在于如何构建矩阵

#include <stdio.h>#include <string.h>#include <stdlib.h>#define le 11typedef struct{    int map[le][le];}re;re init,unit,ar;int k,m;void initdata()//*********************初始化矩阵**********{    int i,j;    for(i = 0;i < 10;i++)        for(j = 0;j < 10;j++){            if(i==j+1)  init.map[i][j] = 1;  //init 矩阵i==j+1时为* 1 *,其余为** 0 **；（后面再输入第一行的值）            else       init.map[i][j] = 0;            if(i == j)   unit.map[i][j] = 1; //unit 矩阵对角线为** 1 **其余都为** 0 **            else         unit.map[i][j] = 0;            ar.map[i][j] = 0;        }    for(i = 9;i >= 0;i--)        ar.map[9-i][0] = i; //ar 矩阵赋值为i（因为k《=9时等于自身）}//*********************************re mul(re va,re vb)//**********************************{    int i,j,k;    re c;    //re c;    for(i = 0;i < 10;i++)        for(j = 0;j < 10;j++){            c.map[i][j] = 0;            for(k = 0;k < 10;k++)                c.map[i][j] += va.map[i][k] * vb.map[k][j];            c.map[i][j] %= m;        }    return c;}//*************************************************************int cal()//*******************************************************{    int i=k-9;  //k>=10的时候*****    re p = unit,q = init,c; //定义p=unit（1），q=init(输入的_),还有一个新的c    while(i>1){        if(i%2){   p = mul(p,q);  i--; }        else   {   q = mul(q,q);  i/=2; }    }    c = mul(p,q);    c = mul(c,ar);  //最后再乘以（9,8,7,6,5,4,3,2,1）    return c.map[0][0];}//***********************************************************int main(){    initdata();int i,ans;    while(scanf("%d%d",&k,&m)==2){for(i=0;i<10;i++)scanf("%d",&init.map[0][i]);//输入第一行矩阵的值**initif(k <= 9){printf("%d\n",k%m);continue;}ans = cal(); //cal 处理函数**********printf("%d\n",ans);    }    return 0;}