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hanyixuanten
2026-09-16 19:39:01 +08:00
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commit 67f1f4d5cc
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AT_agc013_e.cpp
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#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;
}
solutions/AT_agc013_e.md
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# 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;
}
```