题目
A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6512 Accepted Submission(s): 3996
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
题目链接:戳这里
思路
k的值比较大,所以递推是不行的。
那么就考虑矩阵快速幂。
根据 公式构造矩阵,然后直接模板就ok了。
代码
#include<cstdio>
using namespace std;
const int maxn = 10;
int k,MOD;
#define mod(x) ((x)%MOD)
struct Matrix{
int a[maxn][maxn];
void init(){
for(int i=0;i < maxn;i++){
for(int j=0;j < maxn;j++){
a[i][j] = 0;
}
}
}
};
Matrix mul(Matrix A,Matrix B){
Matrix res;
res.init();
for(int i=0;i < maxn;i++){
for(int j=0;j < maxn;j++){
for(int k=0;k < maxn;k++){
res.a[i][j] = mod(res.a[i][j] + mod(A.a[i][k]*B.a[k][j]));
}
}
}
return res;
}
Matrix poww(Matrix A,Matrix B,int n){
Matrix res = B;
while(n){
if(n&1)res = mul(res,A);
A = mul(A,A);
n >>= 1;
}
return res;
}
void outPut(Matrix A){
for(int i=0;i < maxn;i++){
for(int j=0;j < maxn;j++){
printf("%d ",A.a[i][j]);
}
printf("\n");
}
}
int main(){
while(~scanf("%d%d",&k,&MOD)){
Matrix A,B;
A.init();
B.init();
for(int i=9;i >= 0;i--){
A.a[0][9-i] = i;
}
for(int i=0;i <= 9;i++){
scanf("%d",&B.a[i][0]);
}
for(int i=0;i < 9;i++){
B.a[i][i+1] = 1;
}
A = poww(B,A,k-9);
printf("%d\n",A.a[0][0]);
}
return 0;
}