【ZOJ 3690】 Choosing number (矩阵快速幂)

【ZOJ 3690】 Choosing number (矩阵快速幂)

Choosing number Time Limit: 2 Seconds       Memory Limit: 65536 KB

There are n people standing in a row. And There are m numbers, 1.2...m. Every one should choose a number. But if two persons standing adjacent to each other choose the same number, the number shouldn't equal or less than k. Apart from this rule, there are no more limiting conditions.

And you need to calculate how many ways they can choose the numbers obeying the rule.

Input

There are multiple test cases. Each case contain a line, containing three integer n (2 ≤ n ≤ 10⁸), m (2 ≤ m ≤ 30000), k(0 ≤ k ≤ m).

Output

One line for each case. The number of ways module 1000000007.

Sample Input

4 4 1

Sample Output

216

好久不用矩快 没想到。。

递推公式推出来了 上来就想用dp 结果发现太大开不出 其实当时仔细想想只有i-1有价值 之前的都是占用空间。。 10^8会超时 那就矩快呗。。。背锅

题意很好懂 1~m m个数 n个人进行选择 要求如果有相邻的人选择相同的数 这个数必须大于k

递推公式推导:

我是用dp思维来推的 

dp[i][0]表示第i个人与前一个人选择的数不同 同时该数大于k 

dp[i][1]表示第i个人与前一个人选择的数不同 同时该数小于等于k 

dp[i][2]表示第i个人与前一个人选择的数相同 (强制大于k)

转移方程:

dp[i][0] = dp[i-1][0]*(m-k-1) + dp[i-1][1]*(m-k) + dp[i-1][2]*(m-k-1)

dp[i][1] = dp[i-1][0]*k + dp[i-1][1]*(k-1) + dp[i-1][2]*k

dp[i][2] = dp[i-1][0] + dp[i-1][2]

按照第二维的含义很容易就能得出推导公式 接下来就是找转移矩阵

m-k-1 m-k m-k-1

k k-1 k

1 0 1

要注意k == m时m-k-1->0 k == 0时 k-1->0

然后用矩快就行了

代码如下:

#include <bits/stdc++.h>
#define LL long long

using namespace std;
const int mod = 1e9+7;

class Matrix
{
public:
    LL mt[3][3];
    Matrix()//初始化空矩阵
    {
        memset(mt,0,sizeof(mt));
    }
    Matrix(int opt)//初始化单位矩阵
    {
        memset(mt,0,sizeof(mt));
        mt[0][0] = mt[1][1] = mt[2][2] = 1;
    }
    Matrix(int m,int k)//构造转移矩阵
    {
        mt[0][0] = mt[0][2] = max(m-k-1,0);
        mt[0][1] = m-k;
        mt[1][0] = mt[1][2] = k;
        mt[1][1] = max(k-1,0);
        mt[2][0] = mt[2][2] = 1;
        mt[2][1] = 0;
    }
    Matrix operator *(const Matrix &x2)//矩阵乘法
    {
        Matrix x;
        for(int i = 0; i < 3; ++i)
            for(int j = 0; j < 3; ++j)
                for(int k = 0; k < 3; ++k)
                    x.mt[i][j] += (mt[i][k]*x2.mt[k][j])%mod;

        return x;
    }

    void output()//DeBug
    {
        for(int i = 0; i < 3; ++i)
        {
            for(int j = 0; j < 3; ++j) cout<<mt[i][j]<<' ';
            cout<<endl;
        }
        cout<<"--------------------"<<endl;
    }
};


Matrix pow(Matrix x,int b)//矩阵乘法
{
    Matrix ans = Matrix(1);
    while(b)
    {
        if(b&1) ans = ans*x;
        b >>= 1;
        x = x*x;
       // ans.output();
    }
    return ans;
}

int main()
{
    int n,m,k;
    LL ans;

    while(~scanf("%d %d %d",&n,&m,&k))
    {
        Matrix x = pow(Matrix(m,k),n-1);

        ans = 0;
        for(int i = 0; i < 3; ++i)
        {
            for(int j = 0; j < 2; ++j)
            {
                if(j == 0) ans = (ans+(m-k)*x.mt[i][j])%mod;
                else if(j == 1) ans = (ans+k*x.mt[i][j])%mod;
            }
        }
        printf("%lld\n",ans);
    }

    return 0;
}

PS:发现递推式想麻烦了

直接dp[i][2]就够用

dp[i][0] 表示第i人选<=k的数的方案数

dp[i][1] 表示第i人选 >k 的数的方案数

dp[i][0] = dp[i-1][0]*(k-1)+dp[i-1][1]*k

dp[i][1] = (dp[i-1][0]+dp[i-1][1))*(m-k)

矩阵快速幂部分是大同小异的

代码如下:

#include <bits/stdc++.h>
#define LL long long

using namespace std;
const int mod = 1e9+7;

class Matrix
{
public:
    LL mt[2][2];
    Matrix()
    {
        memset(mt,0,sizeof(mt));
    }
    Matrix(int opt)
    {
        memset(mt,0,sizeof(mt));
        mt[0][0] = mt[1][1] = 1;
    }
    Matrix(int m,int k)
    {
        mt[1][0] = mt[1][1] = m-k;
        mt[0][0] = max(k-1,0);
        mt[0][1] = k;
    }
    Matrix operator *(const Matrix &x2)
    {
        Matrix x;
        for(int i = 0; i < 2; ++i)
            for(int j = 0; j < 2; ++j)
                for(int k = 0; k < 2; ++k)
                    x.mt[i][j] += (mt[i][k]*x2.mt[k][j])%mod;

        return x;
    }

    void output()
    {
        for(int i = 0; i < 3; ++i)
        {
            for(int j = 0; j < 3; ++j) cout<<mt[i][j]<<' ';
            cout<<endl;
        }
        cout<<"--------------------"<<endl;
    }
};


Matrix pow(Matrix x,int b)
{
    Matrix ans = Matrix(1);
    while(b)
    {
        if(b&1) ans = ans*x;
        b >>= 1;
        x = x*x;
       // ans.output();
    }
    return ans;
}

int main()
{
    int n,m,k;
    LL ans;

    while(~scanf("%d %d %d",&n,&m,&k))
    {
        Matrix x = pow(Matrix(m,k),n-1);

        ans = 0;
        for(int i = 0; i < 2; ++i)
        {
            for(int j = 0; j < 2; ++j)
            {
                if(j == 1) ans = (ans+(m-k)*x.mt[i][j])%mod;
                else if(j == 0) ans = (ans+k*x.mt[i][j])%mod;
            }
        }
        printf("%lld\n",ans);
    }

    return 0;
}