AT_agc013_e [AGC013E] Placing Squares 题解

把 $\prod a_i^2$ 转化成组合意义，即在每个正方形的底边中放不同的两个球的方案数。

设 $f_{i,j}$ 表示考虑到第 $i$ 个点，最后一条底边放了 $j$ 个球，

若第 $i$ 个点被标记，则：

$$\begin{array}{rl}& f_{i+1,0}=f_{i,0}\\ & f_{i+1,1}=2f_{i,0}+f_{i,1}\\ & f_{i+1,2}=f_{i,0}+f_{i,1}+f_{i,2}\end{array}$$

若第 $i$ 个点没有被标记，则：

$$\begin{array}{rl}& f_{i+1,0}=f_{i,0}+f_{i,2}\\ & f_{i+1,1}=2f_{i,0}+f_{i,1}+2f_{i,2}\\ & f_{i+1,2}=f_{i,0}+f_{i,1}+2f_{i,2}\end{array}$$

矩阵加速即可。

#include <cstdio>
#include <cstring>
#define M 1000000007
#define int long long
int n, m, a[100050];
struct S
{
    int a[3][3];
    S() { memset(a, 0, sizeof a); }
    S operator*(S b)
    {
        S c;
        for (int i = 0; i < 3; ++i)
            for (int j = 0; j < 3; ++j)
            {
                for (int k = 0; k < 3; ++k)
                    c.a[i][j] += a[i][k] * b.a[k][j];
                c.a[i][j] %= M;
            }
        return c;
    }
} z, x, y;
S P(S x, int y)
{
    S q;
    for (int i = 0; i < 3; ++i)
        q.a[i][i] = 1;
    for (; y; y >>= 1, x = x * x)
        if (y & 1)
            q = q * x;
    return q;
}
signed main()
{
    for (int i = 0; i < 3; ++i)
        z.a[i][i] = 1;
    x.a[0][0] = x.a[0][2] = x.a[1][1] = x.a[1][2] = x.a[2][0] = y.a[0][0] = y.a[0][2] = y.a[1][1] = y.a[1][2] = y.a[2][2] = 1;
    x.a[0][1] = x.a[2][1] = x.a[2][2] = y.a[0][1] = 2;
    scanf("%lld%lld", &n, &m);
    if (!m)
        return !printf("%lld", P(x, n).a[0][2]);
    a[0] = -1;
    for (int i = 1; i <= m; ++i)
        scanf("%d", a + i), z = z * P(x, a[i] - a[i - 1] - 1) * y;
    z = z * P(x, n - a[m] - 1);
    printf("%lld", z.a[0][2]);
    return 0;
}