POJ 3233 Matrix Power Series Matrix Rapid Power + binary summation, poj3233

POJ 3233 Matrix Power Series Matrix Rapid Power + binary summation, poj3233

Matrix fast power, please refer to template http://www.cnblogs.com/pach/p/5978475.html

If sum = A + A2 + a3... + Ak is directly accumulated, it will definitely time out,

Sum = A + A2 +... + Ak/2 + A (k/2) * (A + A2 +... + Ak/2) When k is an even number;

Sum = A + A2 +... + A (k-1)/2 + A (k-1)/2) * (A + A2 +... + A (k-1)/2) + Ak k is an odd number.

Then recursive binary summation

PS: At the beginning, mat defined _ int64, so it contributed n times of TLE...

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int n,m;const int N=55;struct Mat{    int mat[N][N];};Mat Multiply(Mat a, Mat b){    Mat c;    memset(c.mat, 0, sizeof(c.mat));    for(int k = 0; k < n; ++k)        for(int i = 0; i < n; ++i)            if(a.mat[i][k])                for(int j = 0; j < n; ++j)                    if(b.mat[k][j])                        c.mat[i][j] = (c.mat[i][j] +a.mat[i][k] * b.mat[k][j])%m;    return c;}Mat QuickPower(Mat a, int k){    Mat c;    memset(c.mat,0,sizeof(c.mat));    for(int i = 0; i < n; ++i)        c.mat[i][i]=1;    for(; k; k >>= 1)    {        if(k&1) c = Multiply(c,a);        a = Multiply(a,a);    }    return c;}Mat Add(Mat a,Mat b){    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)            a.mat[i][j]=(a.mat[i][j]+b.mat[i][j])%m;    return a;}Mat Solve(Mat a,int k){    if(k==1)        return a;    Mat e,ret;    memset(e.mat,0,sizeof(e.mat));    for(int i=0; i<n; i++)        e.mat[i][i]=1;    ret=Multiply(Add(e,QuickPower(a,k>>1)),Solve(a,k>>1));    if(k%2)        return Add(ret,QuickPower(a,k));    return ret;}int main(){    //freopen("in.txt","r",stdin);    int k;    scanf("%d%d%d",&n,&k,&m);    Mat a;    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)            scanf("%d",&a.mat[i][j]);    Mat ans=Solve(a,k);    for(int i=0; i<n; i++)    {        for(int j=0; j<n-1; j++)            printf("%d ",ans.mat[i][j]);        printf("%d\n",ans.mat[i][n-1]);    }    return 0;}