zoj 3690（递推+矩阵优化）

其实比较简单的一道递推题。当时我却没有什么想法真的是太失败了。。

F[X] 表示前X个人，且最后一个选的是比k大的数的总人数

K[X] 表示前X个人，且最后一个选的是<=k 的总人数

则

F[X]=F[X-1]*(m-k)+K[X-1]*(m-k)

K[X]=F[X-1]*k+K[X]*(k-1)

构造出矩阵求解即可

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

Author: GU, Shenlong
Contest: ZOJ Monthly, March 2013

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

#define MOD 1000000007

typedef long long int LL;


struct node
{
    LL g[2][2];
};


int n,m,k;

node cal(node x,node y)
{
    node tmp;
    tmp.g[0][0]=0; tmp.g[0][1]=0;
    tmp.g[1][0]=0; tmp.g[1][1]=0;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                tmp.g[i][j]+=(x.g[i][k]*y.g[k][j])%MOD;
    return tmp;
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        node tmp;
        tmp.g[0][0]=m-k; tmp.g[0][1]=k;
        tmp.g[1][0]=m-k; tmp.g[1][1]=k-1;
        node ans;
        ans.g[0][0]=1; ans.g[0][1]=0;
        ans.g[1][0]=0; ans.g[1][1]=1;
        n--;
        while(n)
        {
            if(n&1)
                ans=cal(ans,tmp);
            n>>=1;
            tmp=cal(tmp,tmp);
        }
        LL ans1=0,ans2=0;
        ans1=((ans.g[0][0]*(m-k))%MOD+(ans.g[1][0]*k)%MOD)%MOD;
        ans2= ((ans.g[0][1]*(m-k))%MOD+(ans.g[1][1]*k)%MOD)%MOD;
        printf("%lld\n",(ans1+ans2)%MOD);
    }
    return 0;
}