原题http://acm.hdu.edu.cn/showproblem.php?pid=1757
A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2526 Accepted Submission(s): 1469
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
//本题只要把矩阵构造出来，接下来的就是套模板了
//构建的矩阵相乘的类型为AAAAAAB，B为已知的函数前几项
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
using namespace std;
#define max 15
int k,mod;
struct Matrix{
int m[max][max];
};
Matrix unit,init;
void Init(){
int i;
memset(init.m,0,sizeof(init.m));
for(i=1;i<10;i++){
init.m[i][i-1] = 1;//自己构建的那个矩阵
}
memset(unit.m,0,sizeof(unit.m));
for(i=0;i<10;i++){
unit.m[i][i] = 1;
}
}
Matrix Mul(Matrix a,Matrix b){
Matrix c;
int i,j,k;
for(i=0;i<10;i++){
for(j=0;j<10;j++){
c.m[i][j] = 0;
for(k=0;k<10;k++){
c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
}
c.m[i][j]%=mod;
}
}
return c;
}
Matrix Pow(Matrix a,Matrix b,int x){
while(x){
if(x&1){
b = Mul(a,b);
}
a = Mul(a,a);
x>>=1;
}
return b;
}
int main(){
int i;
while(~scanf("%d%d",&k,&mod)){
Init();
for(i=0;i<10;i++){
scanf("%d",&init.m[0][i]);
}
if(k < 10){
printf("%d\n",k%mod);
}
Matrix res = Pow(init,unit,k-9);
int ans = 0;
for(i=0;i<10;i++){
ans+=(res.m[0][i]*(9-i))%mod;//因为这个函数的前10项为9,8,7,6,5,4,3,2,1,0
}
ans%=mod;
printf("%d\n",ans);
}
return 0;
}
HDU1757矩阵的简单运用