E. Okabe and El Psy Kongroo
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to know how many different walks he can take in his city safely. Okabe's city can be represented as all points (x, y) such that x and y are non-negative. Okabe starts at the origin (point (0, 0)), and needs to reach the point (k, 0). If Okabe is currently at the point (x, y), in one step he can go to (x + 1, y + 1), (x + 1, y), or (x + 1, y - 1).

Additionally, there are n horizontal line segments, the i-th of which goes from x = ai to x = bi inclusive, and is at y = ci. It is guaranteed that a1 = 0, an ≤ k ≤ bn, and ai = bi - 1 for 2 ≤ i ≤ n. The i-th line segment forces Okabe to walk with y-value in the range 0 ≤ y ≤ ci when his xvalue satisfies ai ≤ x ≤ bi, or else he might be spied on. This also means he is required to be under two line segments when one segment ends and another begins.

Okabe now wants to know how many walks there are from the origin to the point (k, 0) satisfying these conditions, modulo 109 + 7.

Input

The first line of input contains the integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1018) — the number of segments and the destination xcoordinate.

The next n lines contain three space-separated integers aibi, and ci (0 ≤ ai < bi ≤ 1018, 0 ≤ ci ≤ 15) — the left and right ends of a segment, and its y coordinate.

It is guaranteed that a1 = 0, an ≤ k ≤ bn, and ai = bi - 1 for 2 ≤ i ≤ n.

Output

Print the number of walks satisfying the conditions, modulo 1000000007 (109 + 7).

Examples
input
1 3
0 3 3
output
4
input
2 6
0 3 0
3 10 2
output
4
Note

The graph above corresponds to sample 1. The possible walks are:

The graph above corresponds to sample 2. There is only one walk for Okabe to reach (3, 0). After this, the possible walks are:

题目链接:CF 821E

很好的一道题,用dp[x][y]表示到坐标$(x,y)$的方案数,容易可以想到简单的递推:$dp[x][y]=dp[x-1][y]+dp[x-1][y-1]+dp[x-1][y+1]$,但是由于x太大不能直接写dp,看这个式子,发现当前的dp[x]只跟dp[x-1]有关系,因此实际上只需要两个一维数组就可以完成这种迭代,那如何加速迭代呢?用矩阵快速幂,把dp[x-1][y]放到矩阵A的第一行,dp[x][y]显然是转移之后的矩阵A的第一行,如何转移?y从上一次的y-1,y+1,y进行转移,因此构造中间矩阵B,B[i][j]=i走到j是否可行,然后一条线段一条线段地进行转移,当上一次转移线段的高度为3,当前线段高度为2,那显然3这个高度已经越界了,因此转移之前要把A矩阵的越界位置方案数置0,然后用当前的高度作为行数进行转移,否则会多算,最后注意一下长度不能超过k,不然也会多算

代码:

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 110;
const int M = 16;
const LL mod = 1000000007LL;
int row;
struct Mat
{
LL A[M][M];
void zero()
{
CLR(A, 0);
}
Mat operator*(Mat b)
{
Mat c;
c.zero();
for (int i = 0; i < row; ++i)
{
for (int k = 0; k < row; ++k)
{
if (A[i][k])
{
for (int j = 0; j < row; ++j)
{
if (b.A[k][j])
c.A[i][j] = (c.A[i][j] + A[i][k] * b.A[k][j]) % mod;
}
}
}
}
return c;
}
friend Mat operator^(Mat a, LL b)
{
Mat r;
r.zero();
for (int i = 0; i < row; ++i)
r.A[i][i] = 1;
while (b)
{
if (b & 1)
r = r * a;
a = a * a;
b >>= 1;
}
return r;
}
};
LL a[N], b[N];
int c[N]; int main(void)
{
int n, i, j;
LL k;
while (~scanf("%d%I64d", &n, &k))
{
for (i = 0; i < n; ++i)
scanf("%I64d%I64d%d", &a[i], &b[i], &c[i]);
if (b[n - 1] > k)
b[n - 1] = k;
Mat A,B;
A.zero();
A.A[0][0] = 1;
B.zero();
for (j = 0; j <= 15; ++j) //列
{
B.A[j][j] = 1;
if (j - 1 >= 0)
B.A[j][j - 1] = 1;
if (j + 1 <= 15)
B.A[j][j + 1] = 1;
}
for (i = 0; i < n; ++i)
{
row = c[i] + 1;
for (j = c[i] + 1; j < M; ++j)
A.A[0][j] = 0;
A = A * (B ^ (b[i] - a[i]));
}
printf("%I64d\n", A.A[0][0]);
}
return 0;
}

Codeforces 821E Okabe and El Psy Kongroo(矩阵快速幂)的更多相关文章

  1. Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo 矩阵快速幂优化dp

    E. Okabe and El Psy Kongroo time limit per test 2 seconds memory limit per test 256 megabytes input ...

  2. CF821 E. Okabe and El Psy Kongroo 矩阵快速幂

    LINK 题意:给出$n$条平行于x轴的线段,终点$k$坐标$(k <= 10^{18})$,现在可以在线段之间进行移动,但不能超出两条线段的y坐标所夹范围,问到达终点有几种方案. 思路:刚开始 ...

  3. Codeforces 821E Okabe and El Psy Kongroo

    题意:我们现在位于(0,0)处,目标是走到(K,0)处.每一次我们都可以从(x,y)走到(x+1,y-1)或者(x+1,y)或者(x+1,y+1)三个位子之一.现在一共有N段线段,每条线段都是平行于X ...

  4. codeforces E. Okabe and El Psy Kongroo(dp+矩阵快速幂)

    题目链接:http://codeforces.com/contest/821/problem/E 题意:我们现在位于(0,0)处,目标是走到(K,0)处.每一次我们都可以从(x,y)走到(x+1,y- ...

  5. Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo DP+矩阵快速幂加速

    E. Okabe and El Psy Kongroo     Okabe likes to take walks but knows that spies from the Organization ...

  6. Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo dp+矩阵快速幂

    E. Okabe and El Psy Kongroo   Okabe likes to take walks but knows that spies from the Organization c ...

  7. CF821E 【Okabe and El Psy Kongroo】

    首先我们从最简单的dp开始 \(dp[i][j]=dp[i-1][j]+dp[i-1][j+1]+dp[i-1][j-1]\) 然后这是一个O(NM)的做法,肯定行不通,然后我们考虑使用矩阵加速 \( ...

  8. [codeforces821E]Okabe and El Psy Kongroo

    题意:(0,0)走到(k,0),每一部分有一条线段作为上界,求方案数. 解题关键:dp+矩阵快速幂,盗个图,注意ll 关于那条语句为什么不加也可以,因为我的矩阵C,就是因为多传了了len的原因,其他位 ...

  9. Educational Codeforces Round 13 D. Iterated Linear Function (矩阵快速幂)

    题目链接:http://codeforces.com/problemset/problem/678/D 简单的矩阵快速幂模版题 矩阵是这样的: #include <bits/stdc++.h&g ...

随机推荐

  1. PHP无法用下标访问

    php数组分为普通数组和关联数组,普通数组可以用下标访问,而关联数组不可以.

  2. Flask之endpoint错误View function mapping is overwriting an existing endpoint function: ***

    最近在学习Flask, 其中遇到了一个错误, 发现这个问题和Flask, 路由有关系, 所以就记了下来 错误代码: from flask import Flask, render_template, ...

  3. maven-认识

    1.认识maven maven是强大的项目构建工具,也是依赖管理工具 使用maven前提是安装JDK maven非常重要配置文件:setting.xml 3.maven工程 maven工程的约束: 主 ...

  4. Laravel-初步使用

    一.Laravel环境搭建 1.window环境下环境搭建请参考以下链接: 开发环境搭建 - Windows | <Laravel 开发环境部署> | PHP / Laravel 社区文档 ...

  5. python之打印九九乘法表

    配置环境:python 3.6 python编辑器:pycharm 整理成代码如下: #!/usr/bin/env python #-*- coding: utf-8 -*- #九九乘法表 #分析:九 ...

  6. Python学习 :六个标准数据类型

    一.Numbers(数字类型) 数字类型主要分为两种—— 整数(Integer)与 浮点数(Float) 整数分为整型和长整型(在Python3中已经不再区分为整型与长整型,统一称为整型) 注意:数字 ...

  7. Scrapy核心组件

    • 引擎(Scrapy)用来处理整个系统的数据流处理, 触发事务(框架核心) • 调度器(Scheduler)用来接受引擎发过来的请求, 压入队列中, 并在引擎再次请求的时候返回. 可以想像成一个UR ...

  8. python2.7练习小例子(十二)

        12):题目:打印出所有的"水仙花数",所谓"水仙花数"是指一个三位数,其各位数字立方和等于该数本身.例如:153是一个"水仙花数" ...

  9. WPF中的ControlTemplate(控件模板)

    原文:WPF中的ControlTemplate(控件模板) WPF中的ControlTemplate(控件模板)                                             ...

  10. Android 数据库 ANR的例子

    android 开启事务之后,在其他线程是不能进行增删改查操作的.例子如下: 首先,一个线程里面去开启事务,里面对数据库的任何操作都没有. DBAdapter.getInstance().beginT ...