bzoj 4688: One-Dimensional 矩阵乘法

       矩乘sb题。

AC代码如下：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,m,ta,tb,tc,mod;struct matrix{ int p[55][55]; }a,b;matrix operator *(const matrix &x,const matrix &y){matrix z; memset(z.p,0,sizeof(z.p));int i,j,k;for (i=1; i<=n; i++)for (j=1; j<=n; j++)for (k=1; k<=n; k++) z.p[i][k]+=x.p[i][j]*y.p[j][k];for (i=1; i<=n; i++)for (j=1; j<=n; j++) z.p[i][j]%=mod;return z;}int main(){while (~scanf("%d%d%d%d%d%d",&n,&mod,&ta,&tb,&tc,&m) && n){int i;for (i=1; i<=n; i++) scanf("%d",&a.p[1][i]);memset(b.p,0,sizeof(b.p));for (i=1; i<=n; i++){if (i<n) b.p[i][i+1]=ta;b.p[i][i]=tb;if (i>1) b.p[i][i-1]=tc;}for (; m; m>>=1,b=b*b) if (m&1) a=a*b;for (i=1; i<=n; i++) printf("%d%c",a.p[1][i],(i<n)?' ':'\n');}return 0;}

by lych

2016.9.4