【题目链接】:http://codeforces.com/problemset/problem/821/E

【题意】



一开始位于(0,0)的位置;

然后你每次可以往右上,右，右下3走一步;

(x+1,y+1),(x+1,y),(x+1,y-1)

然后有n条横线,限制你在横线与x轴之间;

问你从(0,0)到(k,0)有多少种行走方案;

k可以很大;

保证横线是连续的,即一条横线完,马上接另外一条横线;

【题解】



如果不考虑那么大的横坐标的话;

用最简单的DP

设f[i][j]表示到(i,j)这个点的方案数;

则有f[i][j] = f[i-1][j+1] + f[i-1][j] + f[i-1][j-1];

在这个DP的基础上,写一个矩阵优化;

构造这么一个矩阵

    (1 1 0    ....0)

    (1 1 1 0   ...0)

    (0 1 1 1 0 ...0)

    (0 0 1 1 1 ...0)

    ....

    (0 0 0 0 0...1 1)

    c[i]+1行，c[i]+1列

然后用一个一列的矩阵

f[1]

f[2]

f[3]

...

f[c[i]+1]

去左乘它;

这里f[i]表示当前这一段横线;

到达纵坐标为i的方案有多少种;

把那个矩阵的min(k,b[i])-a[i]次幂求出来；

然后用那个f矩阵去左乘它得到新的f矩阵就好;

每段都这样做;

最后输出f[1];



【Number Of WA】



0



【反思】



没仔细去想矩阵的构造方法,就跳过去了;

还是不敢去做题啊。被吓到了。



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 110;

const int G = 16;       //矩阵大小

const LL MOD = 1e9 + 7;    //模数

struct MX

{

    int v[G+5][G+5];

    void O() { ms(v, 0); }

    void E() { ms(v, 0); for (int i = 1; i <= G; ++i)v[i][i] = 1; }

    void P()

    {

        for (int i = 1; i <= G; ++i)

        {

            for (int j = 1; j <= G; ++j)printf("%d ", v[i][j]); puts("");

        }

    }

    MX operator * (const MX &b) const

    {

        MX c; c.O();

        for (int k = 1; k <= G; ++k)

        {

            for (int i = 1; i <= G; ++i) if (v[i][k])

            {

                for (int j = 1; j <= G; ++j)

                {

                    c.v[i][j] = (c.v[i][j] + (LL)v[i][k] * b.v[k][j]) % MOD;

                }

            }

        }

        return c;

    }

    MX operator + (const MX &b) const

    {

        MX c; c.O();

        for (int i = 1; i <= G; ++i)

        {

            for (int j = 1; j <= G; ++j)

            {

                c.v[i][j] = (v[i][j] + b.v[i][j]) % MOD;

            }

        }

        return c;

    }

    MX operator ^ (LL p) const

    {

        MX y; y.E();

        MX x; memcpy(x.v, v, sizeof(v));

        while (p)

        {

            if (p&1) y = y*x;

            x = x*x;

            p>>=1;

        }

        return y;

    }

}xishu;

struct abc{

    LL a,b;

    int c;

};

int n;

LL k;

LL f[20],nf[20];

abc a[N];

int main(){

    //Open();

    Close();

    cin >> n >> k;

    rep1(i,1,n){

        cin >> a[i].a >> a[i].b >> a[i].c;

    }

    f[1] = 1;

    rep1(i,1,n){

        xishu.O();

        rep1(j,1,a[i].c+1){

            int now =j-1;

            for (int jj = 1;jj <= 3;jj++){

                xishu.v[j][now++] = 1;

            }

        }

        xishu = xishu^(min(k,a[i].b)-a[i].a);

        rep1(j,1,16) nf[j] = 0;

        rep1(j,1,a[i].c+1){

            rep1(k,1,a[i].c+1){

                nf[j] = (nf[j]+f[k]*xishu.v[j][k])%MOD;

            }

        }

        rep1(j,1,16) f[j] = nf[j];

    }

    cout << f[1] << endl;

    return 0;

}