CodeForces 385E Bear in the Field（矩阵快速幂）

这道题主要是有两个值要输出，建立矩阵转移方程还是参考了题解~

对套路的理解还是不够深啊

主要要清楚有几个变量，以及他们之间的关系，这样就好做了~

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

#define LL long long

LL n,sx,sy,dx,dy,t;

struct matrix {
LL m[8][8];
matrix() {
memset(m,0,sizeof(m));
}
};

matrix mul(matrix a,matrix b) {
matrix tmp;
for(int i = 1;i <= 6;i++)
for(int k = 1;k <= 6;k++)
if(a.m[i][k])
for(int j = 1;j <= 6;j++)
tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j] + n) % n;
return tmp;
}

matrix powmul(matrix a,LL t) {
matrix tmp;
for(int i = 1;i <= 6;i++)
tmp.m[i][i] = 1;
while(t) {
if(t & 1)
tmp = mul(tmp,a);
a = mul(a,a);
t >>= 1;
}
return tmp;
}

int main() {
scanf("%lld%lld%lld%lld%lld%lld",&n,&sx,&sy,&dx,&dy,&t);
matrix base;
base.m[1][2] = base.m[1][3] = base.m[1][5] = 1;
base.m[2][1] = base.m[2][4] = base.m[2][5] = 1;
base.m[3][1] = base.m[3][2] = base.m[3][3] = base
4000
.m[3][5] = 1;
base.m[4][1] = base.m[4][2] = base.m[4][4] = base.m[4][5] = 1;
base.m[5][5] = base.m[5][6] = base.m[6][6] = 1;
base.m[2][2] = base.m[1][1] = base.m[1][6] = base.m[2][6] = base.m[3][6] = base.m[4][6] = 2;
/*for(int i = 1;i <= 6;i++)
for(int j = 1;j <= 6;j++)
printf("%lld%c",base.m[i][j],j == 6 ? '\n' : ' ');
cout<<endl;*/
base = powmul(base,t);
/*for(int i = 1;i <= 6;i++)
for(int j = 1;j <= 6;j++)
printf("%lld%c",base.m[i][j],j == 6 ? '\n' : ' ');
cout<<endl;*/
matrix ans;
ans.m[1][1] = sx-1;
ans.m[2][1] = sy-1;
ans.m[3][1] = (dx % n + n) % n;
ans.m[4][1] = (dy % n + n) % n; //含负数要加n再余n
ans.m[6][1] = 1;
ans = mul(base,ans);
printf("%lld %lld\n",ans.m[1][1] + 1,ans.m[2][1] + 1);
}