From 67f1f4d5ccb0d895e96b553661c750231f137d32 Mon Sep 17 00:00:00 2001
From: hanyixuanten <105723997+hanyixuanten@users.noreply.github.com>
Date: Wed, 16 Sep 2026 19:39:01 +0800
Subject: [PATCH] new file:   AT_agc013_e.cpp

---
 AT_agc013_e.cpp          |  98 +++++++++++++++
 solutions/AT_agc013_e.md | 266 +++++++++++++++++++++++++++++++++++++++
 2 files changed, 364 insertions(+)
 create mode 100644 AT_agc013_e.cpp
 create mode 100644 solutions/AT_agc013_e.md

diff --git a/AT_agc013_e.cpp b/AT_agc013_e.cpp
new file mode 100644
index 0000000..1cb62cf
--- /dev/null
+++ b/AT_agc013_e.cpp
@@ -0,0 +1,98 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int mod = 1e9 + 7;
+int n, m;
+int a[100005];
+const long long fir[4][2] = {{0, 0}, {0, 1}, {0, 2}, {0, 1}};
+const long long yes[4][4] = {{0, 0, 0, 0}, {0, 1, 0, 0}, {0, 2, 1, 0}, {0, 1, 1, 1}};
+const long long no[4][4] = {{0, 0, 0, 0}, {0, 1, 0, 1}, {0, 2, 1, 2}, {0, 1, 1, 2}};
+struct Mat
+{
+    long long a[5][5] = {};
+    int n, m;
+    Mat(int _n = 0, int _m = 0, int cpy = 0) : n(_n), m(_m)
+    {
+        if (cpy == 1)
+            for (int i = 1; i <= 3; ++i)
+                a[i][1] = fir[i][1];
+        else if (cpy == 2)
+            for (int i = 1; i <= 3; ++i)
+                for (int j = 1; j <= 3; ++j)
+                    a[i][j] = yes[i][j];
+        else if (cpy == 3)
+            for (int i = 1; i <= 3; ++i)
+                for (int j = 1; j <= 3; ++j)
+                    a[i][j] = no[i][j];
+    }
+    void operator*=(const Mat &b)
+    {
+        if (m != b.n)
+            exit(-1);
+        long long res[5][5] = {};
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                for (int k = 1; k <= m; ++k)
+                    res[i][j] = (res[i][j] + a[i][k] * b.a[k][j]) % mod;
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                a[i][j] = res[i][j];
+        m = b.m;
+    }
+    Mat operator*(const Mat &b)
+    {
+        if (m != b.n)
+            exit(-1);
+        Mat res(n, b.m);
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                for (int k = 1; k <= m; ++k)
+                    res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]) % mod;
+        return res;
+    }
+};
+
+Mat qpow(Mat a, long long k)
+{
+    Mat res(a.n, a.n);
+    for (int i = 1; i <= a.n; ++i)
+        res.a[i][i] = 1;
+    while (k)
+    {
+        if (k & 1)
+            res *= a;
+        a *= a;
+        k >>= 1;
+    }
+    return res;
+}
+Mat yess(3, 3, 2);
+Mat noo(3, 3, 3);
+
+int main()
+{
+    scanf("%d%d", &n, &m);
+    for (int i = 1; i <= m; ++i)
+        scanf("%d", &a[i]);
+    Mat now(3, 1, 1);
+    int pos = 1;
+    for (int i = 1; i <= m; ++i)
+    {
+        int x = a[i];
+        if (x < pos)
+            continue;
+        if (x > pos)
+        {
+            now = qpow(noo, x - pos) * now;
+        }
+        {
+            now = yess * now;
+        }
+        pos = x + 1;
+    }
+    if (n - pos > 0)
+    {
+        now = qpow(noo, n - pos) * now;
+    }
+    printf("%lld\n", now.a[3][1]);
+    return 0;
+}
\ No newline at end of file
diff --git a/solutions/AT_agc013_e.md b/solutions/AT_agc013_e.md
new file mode 100644
index 0000000..db8be77
--- /dev/null
+++ b/solutions/AT_agc013_e.md
@@ -0,0 +1,266 @@
+# AT_agc013_e
+
+[luoshunran](https://www.acwing.com/file_system/file/content/whole/index/content/14321537/)
+
+## 20 pts O(n)
+
+把正方形的交界处看作隔板，在每两个隔板之间放入一黑一白两个球，正方形的大小也就等于区间内放入两个球的方案数，正方形的面积的乘积就是所有的放球方案乘法原理乘起来。于是变成了求有多少种放隔板、放球的方案。
+
+做 dp $f(i, 0/1/2)$，代表目前这个区间内放入了几个球的方案。
+
+注意 $f(1, 1) = 2$
+
+若 $i - i + 1$ 之间可以放隔板，
+
+- $f(i+1, 0) = f(i, 0) + f(i, 2)$
+- $ f(i+1, 1) = 2 * f(i, 0) + f(i, 1) + 2 * f(i, 2)$, 这里乘以 2 代表决策放入的是黑球还是白球
+- $ f(i+1, 2) = f(i, 0) + f(i, 1) + 2 * f(i, 2) $
+
+类似地，也可以列出不可以放隔板的情况。
+
+- $f(i+1, 0) = f(i, 0)$
+- $f(i+1, 1) = 2 * f(i, 0) + f(i, 1)$
+- $f(i+1, 2) = f(i, 0) + f(i, 1) + f(i, 2)$
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+const int mod = 1e9 + 7;
+int n, m;
+long long f[5000005][3];
+int a[100005];
+int main()
+{
+    scanf("%d%d", &n, &m);
+    for (int i = 1; i <= m; ++i)
+        scanf("%d", &a[i]);
+    sort(a + 1, a + m + 1);
+    int tot = 1;
+    f[1][0] = 1, f[1][1] = 2, f[1][2] = 1;
+    for (int i = 1; i < n; ++i)
+    {
+        if (i == a[tot])
+        {
+            tot++;
+            f[i + 1][0] = f[i][0];
+            f[i + 1][1] = (f[i][0] * 2 + f[i][1]) % mod;
+            f[i + 1][2] = (f[i][0] + f[i][1] + f[i][2]) % mod;
+        }
+        else
+        {
+            f[i + 1][0] = (f[i][0] + f[i][2]) % mod;
+            f[i + 1][1] = (f[i][0] * 2 + f[i][1] + 2 * f[i][2]) % mod;
+            f[i + 1][2] = (f[i][0] + f[i][1] + f[i][2] * 2) % mod;
+        }
+    }
+    printf("%lld\n", f[n][2]);
+    return 0;
+}
+```
+
+## 90 pts O(m log n)
+
+注意到一个区间内需要多次转移，非常浪费。可以考虑使用矩阵快速幂优化。
+
+对于非标记点，转移矩阵为：
+
+$$
+T=\begin{bmatrix}
+    1&0&1\\
+    2&1&2\\
+    1&1&2
+\end{bmatrix}
+$$
+
+对于标记点：
+
+$$
+T=\begin{bmatrix}
+    1&0&0\\
+    2&1&0\\
+    1&1&1
+\end{bmatrix}
+$$
+
+初始位置矩阵为：
+
+$$
+T=\begin{bmatrix}
+    1\\
+    2\\
+    1
+\end{bmatrix}
+$$
+
+矩阵快速幂代码可参考[P1962](https://luogu.com.cn/problem/P1962)
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+const long long MOD = 1000000007LL;
+struct Mat {
+    long long a[105][105];
+    int n, m;
+    Mat(int _n = 0, int _m = 0) : n(_n), m(_m) {
+        memset(a, 0, sizeof(a));
+    }
+    void operator*=(const Mat &b) {
+        if (m != b.n) {
+            exit(-1);
+        }
+        long long res[105][105] = {};
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j <= b.m; ++j) {
+                for (int k = 1; k <= m; ++k) {
+                    res[i][j] = (res[i][j] + a[i][k] * b.a[k][j]) % MOD;
+                }
+            }
+        }
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j <= b.m; ++j) {
+                a[i][j] = res[i][j];
+            }
+        }
+        m = b.m;
+    }
+};
+
+Mat qpow(Mat a, long long k) {
+    Mat res(a.n, a.n);
+    for (int i = 1; i <= a.n; ++i) {
+        res.a[i][i] = 1;
+    }
+    while (k) {
+        if (k & 1) res *= a;
+        a *= a;
+        k >>= 1;
+    }
+    return res;
+}
+
+int main() {
+    long long n;
+    scanf("%lld", &n);
+
+    if (n <= 2) {
+        printf("1\n");
+        return 0;
+    }
+
+    Mat base1(1, 2), base2(2, 2);
+
+    base1.a[1][1] = 1;
+    base1.a[1][2] = 1;
+    base2.a[1][1] = 1;
+    base2.a[1][2] = 1;
+    base2.a[2][1] = 1;
+    base2.a[2][2] = 0;
+    base2 = qpow(base2, n - 2);
+    base1 *= base2;
+    printf("%lld\n", base1.a[1][1]);
+    return 0;
+}
+```
+
+最终程序：
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+const int mod = 1e9 + 7;
+int n, m;
+int a[100005];
+const long long fir[4][2] = {{0, 0}, {0, 1}, {0, 2}, {0, 1}};
+const long long yes[4][4] = {{0, 0, 0, 0}, {0, 1, 0, 0}, {0, 2, 1, 0}, {0, 1, 1, 1}};
+const long long no[4][4] = {{0, 0, 0, 0}, {0, 1, 0, 1}, {0, 2, 1, 2}, {0, 1, 1, 2}};
+struct Mat
+{
+    long long a[5][5] = {};
+    int n, m;
+    Mat(int _n = 0, int _m = 0, int cpy = 0) : n(_n), m(_m)
+    {
+        if (cpy == 1)
+            for (int i = 1; i <= 3; ++i)
+                a[i][1] = fir[i][1];
+        else if (cpy == 2)
+            for (int i = 1; i <= 3; ++i)
+                for (int j = 1; j <= 3; ++j)
+                    a[i][j] = yes[i][j];
+        else if (cpy == 3)
+            for (int i = 1; i <= 3; ++i)
+                for (int j = 1; j <= 3; ++j)
+                    a[i][j] = no[i][j];
+    }
+    void operator*=(const Mat &b)
+    {
+        if (m != b.n)
+            exit(-1);
+        long long res[5][5] = {};
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                for (int k = 1; k <= m; ++k)
+                    res[i][j] = (res[i][j] + a[i][k] * b.a[k][j]) % mod;
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                a[i][j] = res[i][j];
+        m = b.m;
+    }
+    Mat operator*(const Mat &b)
+    {
+        if (m != b.n)
+            exit(-1);
+        Mat res(n, b.m);
+        for (int i = 1; i <= n; ++i)
+            for (int j = 1; j <= b.m; ++j)
+                for (int k = 1; k <= m; ++k)
+                    res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]) % mod;
+        return res;
+    }
+};
+
+Mat qpow(Mat a, long long k)
+{
+    Mat res(a.n, a.n);
+    for (int i = 1; i <= a.n; ++i)
+        res.a[i][i] = 1;
+    while (k)
+    {
+        if (k & 1)
+            res *= a;
+        a *= a;
+        k >>= 1;
+    }
+    return res;
+}
+Mat yess(3, 3, 2);
+Mat noo(3, 3, 3);
+
+int main()
+{
+    scanf("%d%d", &n, &m);
+    for (int i = 1; i <= m; ++i)
+        scanf("%d", &a[i]);
+    Mat now(3, 1, 1);
+    int pos = 1;
+    for (int i = 1; i <= m; ++i)
+    {
+        int x = a[i];
+        if (x < pos)
+            continue;
+        if (x > pos)
+        {
+            now = qpow(noo, x - pos) * now;
+        }
+        {
+            now = yess * now;
+        }
+        pos = x + 1;
+    }
+    if (n - pos > 0)
+    {
+        now = qpow(noo, n - pos) * now;
+    }
+    printf("%lld\n", now.a[3][1]);
+    return 0;
+}
+```