HDU-1757--A Simple Math Problem（矩阵乘法）

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

Author
linle
Source
2007省赛集训队练习赛（6）_linle专场
Recommend
lcy
矩阵乘法



如图所示，可以将递推式写成矩阵形式
类似于将斐波那契数列写成矩阵形式



因为数据量很大，需要用到矩阵快速幂
矩阵快速幂见这篇博客
http://www.cnblogs.com/yan-boy/archive/2012/11/29/2795294.html
需要注意的一个小细节是，在进行乘法（不是矩阵乘法）运算时，注意要模上一个数，防止溢出（因为这个WA了好几发）
代码：

#include<bits/stdc++.h>
using namespace std;
struct  node
{
int m[10][10];
node(){
memset(m,0,sizeof(m));
//        for(int i=0;i<10;i++){
//            for(int j=0;j<10;j++)
//                m[i][j]=0;
//        }
}
};
int mod;
node multi(node &a,node &b)
{
node tmp;
for(int i=0;i<10;i++){
for(int j=0;j<10;j++){
int sum=0;
for(int k=0;k<10;k++){
sum+=(a.m[i][k]%mod)*(b.m[k][j]%mod);
}
tmp.m[i][j]=sum;
}
}
return tmp;
}
void e_mat(node &a)
{
for(int i=0;i<10;i++){
a.m[i][i]=1;
}
}
node quick_mul(node &a,int n)
{
node tmp=a;
//e_mat(tmp);
node res;
e_mat(res);
if(n&1){
res=a;
}
n=n>>1;
while(n!=0){
tmp=multi(tmp,tmp);
if(n&1){
res=multi(res,tmp);
}
n=n>>1;
}
return res;
}
void print(node &a)
{
cout<<"---------------------------"<<endl;
for(int i=0;i<10;i++){
cout<<i<<"ÐÐ"<<"\t";
for(int j=0;j<10;j++){
cout<<a.m[i][j]<<"\t";
}
cout<<endl;
}
cout<<"---------------------------"<<endl;
}
int main()
{
//freopen("data.in","r",stdin);
int n;

int ai[11];
node x;
for(int i=0;i<10;i++){
x.m[i][0]=9-i;
}
while(~scanf("%d%d",&n,&mod)){
for(int i=0;i<10;i++){
scanf("%d",ai+i);
}
if(n<10){
printf("%d\n",n);
continue;
}
node a;
//print(a);
for(int i=0;i<10;i++){
a.m[0][i]=ai[i];
}
//print(a);
for(int i=1;i<10;i++){
a.m[i][i-1]=1;
}
//print(a);
node res=quick_mul(a,n-9);
//print(res);
//print(x);
res=multi(res,x);
//print(res);
printf("%d\n",(res.m[0][0])%mod);
}
}