传送门

1.模拟

easy

#include <cstdio>
#define N 500001

int n, m;
int X
, Y
, x
, y
, a = 1, b = 1, p, q;
char s
[1];

int main()
{
int i;
scanf("%d %d", &n, &m);
for(i = 1; i <= n; i++) scanf("%d %d", &X[i], &Y[i]);
for(i = 1; i <= m; i++)
{
scanf("%s", s[i]);
if(s[i][0] == 'm') scanf("%d %d", &x[i], &y[i]);
}
for(i = m; i >= 1; i--)
if(s[i][0] == 'm')
{
p += x[i];
q += y[i];
}
else if(s[i][0] == 'x')
{
a *= -1;
p *= -1;
}
else
{
b *= -1;
q *= -1;
}
for(i = 1; i <= n; i++) printf("%d %d\n", X[i] * a + p, Y[i] * b + q);
return 0;
}

2.矩阵优化

详见 Matrix67

#include <cstdio>
#include <cstring>
#define N 500001

int n, m;
int X
, Y
, x
, y
;
char s
[2];

struct Matrix
{
int n, m;
int a[11][11];
Matrix()
{
n = m = 0;
memset(a, 0, sizeof(a));
}
}sum;

inline Matrix operator * (const Matrix x, const Matrix y)
{
Matrix ret;
ret.n = x.n;
ret.m = y.m;
int i, j, k;
for(i = 1; i <= x.n; i++)
for(j = 1; j <= y.m; j++)
for(k = 1; k <= y.n; k++)
ret.a[i][j] += x.a[i][k] * y.a[k][j];
return ret;
}

int main()
{
int i;
Matrix t;
scanf("%d %d", &n, &m);
sum.n = sum.m = 3;
sum.a[1][1] = sum.a[2][2] = sum.a[3][3] = 1;
for(i = 1; i <= n; i++) scanf("%d %d", &X[i], &Y[i]);
for(i = 1; i <= m; i++)
{
scanf("%s", s[i]);
if(s[i][0] == 'm') scanf("%d %d", &x[i], &y[i]);
}
for(i = m; i >= 1; i--)
{
memset(t.a, 0, sizeof(t.a));
t.n = t.m = 3;
t.a[1][1] = t.a[2][2] = t.a[3][3] = 1;
if(s[i][0] == 'm') t.a[1][3] = x[i], t.a[2][3] = y[i];
if(s[i][0] == 'x') t.a[1][1] = -1;
if(s[i][0] == 'y') t.a[2][2] = -1;
sum = t * sum;
}
for(i = 1; i <= n; i++)
{
memset(t.a, 0, sizeof(t.a));
t.n = 3;
t.m = t.a[3][1] = 1;
t.a[1][1] = X[i];
t.a[2][1] = Y[i];
t = sum * t;
printf("%d %d\n", t.a[1][1], t.a[2][1]);
}
return 0;
}