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+# P16055 [CSPro 31] 坐标变换（其二） 题解
+
+这篇题解主要针对不懂得矩阵乘法的人。主要思想：
+
+视每次操作为：
+
+$$
+newx = x * k1 - y * k2\\
+newy = x * k2 + y * k1\\
+x = newx, y = newy
+$$
+
+其中：
+
+- 操作为 $1$ 时，$k1 = k, k2 = 0$。
+- 操作为 $2$ 时，$k1 = \cos(\theta), k2 = \sin(\theta)$。
+
+## 80 pts 暴力
+
+注意到 $80%$ 的测试点满足 $n, m \le 1000$，此时最大执行次数为 $1000 \times 1000 = 10^6$，可以获得此部分分。
+
+::::info[[80 pts 代码](https://www.luogu.com.cn/record/303033043)]
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int n, m;
+vector<pair<int, double>> o;
+double k1[100005], k2[100005];
+int main() {
+    scanf("%d%d", &n, &m);
+    o.push_back({0, 0});
+    for (int i = 1; i <= n; ++i) {
+        int op;
+        double k;
+        scanf("%d%lf", &op, &k);
+        o.push_back({op, k});
+        if (op == 2) {
+            k1[i] = cos(k);
+            k2[i] = sin(k);
+        } else {
+            k1[i] = k;
+            k2[i] = 0;
+        }
+    }
+    for (int i = 1; i <= m; ++i) {
+        int l, r, x, y;
+        scanf("%d%d%d%d", &l, &r, &x, &y);
+        double nx = x, ny = y;
+        for (int j = l; j <= r; ++j) {
+            if (o[j].first == 1)
+                nx *= o[j].second, ny *= o[j].second;
+            else {
+                double tmpx = nx * k1[j] - ny * k2[j],
+                       tmpy = nx * k2[j] + ny * k1[j];
+                nx = tmpx, ny = tmpy;
+            }
+        }
+        printf("%f %f\n", nx, ny);
+    }
+    return 0;
+}
+```
+
+::::
+
+## 100 pts st 表
+
+注意到连续一段的操作可以由各部分操作叠加，可使用 st 表解决。
+
+::::info[[100pts 代码](https://www.luogu.com.cn/record/303042049)]
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int n, m;
+vector<pair<int, double>> o;
+double st1[200005][21], st2[200005][21];
+int main() {
+    scanf("%d%d", &n, &m);
+    o.push_back({0, 0});
+    for (int i = 1; i <= n; ++i) {
+        int op;
+        double k;
+        scanf("%d%lf", &op, &k);
+        o.push_back({op, k});
+        if (op == 1) {
+            st1[i][0] = k;
+            st2[i][0] = 0;
+
+        } else {
+            st1[i][0] = cos(k);
+            st2[i][0] = sin(k);
+        }
+    }
+    for (int i = 1; i <= 20; i++) {
+        for (int j = 1; j + (1 << i) - 1 <= n; j++) {
+            double tmp11 = st1[j][i - 1], tmp12 = st2[j][i - 1];
+            double tmp21 = st1[j + (1 << (i - 1))][i - 1],
+                tmp22 = st2[j + (1 << (i - 1))][i - 1];
+            st1[j][i] = tmp11 * tmp21 - tmp12 * tmp22;
+            st2[j][i] = tmp11 * tmp22 + tmp12 * tmp21;
+        }
+    }
+    while (m--) {
+        int l, r, x, y;
+        scanf("%d%d%d%d", &l, &r, &x, &y);
+        double nx = x, ny = y;
+        r++;
+        for (int i = 20; i >= 0; i--) {
+            if (l + (1 << i) <= r) {
+                double tmpx = st1[l][i] * nx - st2[l][i] * ny,
+                       tmpy = st2[l][i] * nx + st1[l][i] * ny;
+                nx = tmpx, ny = tmpy;
+                l += 1 << i;
+            }
+        }
+        printf("%f %f\n", nx, ny);
+    }
+    return 0;
+}
+```
+
+::::