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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
题解:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int maxn=1005;
int k,m;
struct Matrix{
int mat[10][10];
}ans,res;
Matrix Mul(Matrix a,Matrix b){//矩阵乘法
Matrix tmp;
for(int i=0;i<10;i++){
for(int j=0;j<10;j++)
tmp.mat[i][j]=0;
}
for(int i=0;i<10;i++){
for(int j=0;j<10;j++){
for(int h=0;h<10;h++){
tmp.mat[i][j]=(tmp.mat[i][j]+a.mat[i][h]*b.mat[h][j])%m;
}
}
}
return tmp;
}
void quick_pow(int n){//快速幂
for(int i=0;i<10;i++){
for(int j=0;j<10;j++){
if(i==j) ans.mat[i][j]=1;
else ans.mat[i][j]=0;
}
}
while(n){
if(n&1)
ans=Mul(ans,res);
res=Mul(res,res);
n>>=1;
}
}
int main(){
while(~scanf("%d%d",&k,&m)){
if(k<10){
printf("%d\n",k);
continue;
}
memset(res.mat,0,sizeof res.mat);
for(int i=0;i<10;i++)
scanf("%d",&res.mat[0][i]);
for(int i=1;i<10;i++)
res.mat[i][i-1]=1;
quick_pow(k-9);
int sum=0;
for(int i=0;i<10;i++)
sum+=ans.mat[0][i]*(9-i);
printf("%d\n",sum%m);
}
return 0;
}