HDU 1757 a simple math problem (matrix fast power)

Address: HDU 1757

Finally, the matrix will be constructed. In fact, it is not difficult, just blame yourself for being stupid .. =!

F (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10)
The constructed matrix is: (The Matrix constructed in my code is reversed up and down)
| 0 1 0... 0 | f0 | F1 |
| 0 0 1 0... 0 | F1 | F2 |
| ...... 1 | * |... | = |
| A9 A8 ...... A0 | F9 | F10 |

Then, according to the combination Law of the matrix, we can first calculate the (K-9) Power of the constructed matrix. Finally, calculate the first number.

The Code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;int mod, a[20];struct matrix{    int ma[20][20];} init, res, s;matrix Mult(matrix x, matrix y){    int i, j, k;    matrix tmp;    for(i=0; i<10; i++)    {        for(j=0; j<10; j++)        {            tmp.ma[i][j]=0;            for(k=0; k<10; k++)            {                tmp.ma[i][j]=(tmp.ma[i][j]+x.ma[i][k]*y.ma[k][j])%mod;            }        }    }    return tmp;}matrix Pow(matrix x, int k){    matrix tmp;    int i, j;    for(i=0; i<10; i++) for(j=0; j<10; j++) tmp.ma[i][j]=(i==j);    while(k)    {        if(k&1) tmp=Mult(tmp,x);        x=Mult(x,x);        k>>=1;    }    return tmp;}int main(){    int k, x, i, j;    while(scanf("%d%d",&k,&mod)!=EOF)    {        if(k<10)        {            printf("%d\n",k%mod);            continue ;        }        for(i=9; i>=0; i--)        {            a[i]=9-i;        }        for(i=0; i<10; i++)        {            scanf("%d",&x);            init.ma[0][i]=x%mod;        }        for(i=1; i<10; i++)        {            for(j=0; j<10; j++)            {                init.ma[i][j]=(i==j+1);            }        }        res=Pow(init,k-9);        /*for(i=0; i<10; i++)        {            for(j=0; j<10; j++)            {                printf("%d ",res.ma[i][j]);            }            puts("");        }*/        int ans=0;        for(j=0; j<10; j++)        {            ans=(ans+res.ma[0][j]*a[j])%mod;            //printf("%d %d %d\n",res.ma[i][j],a[i],ans);        }        printf("%d\n",ans);    }    return 0;}

HDU 1757 a simple math problem (matrix fast power)