BZOJ 4005 [Jloi 2015] lie to me.

First, we can get: the number of each row is different, so each row will have and only one in $[0, m]$ number does not appear.

We can consider setting the $Dp [i][j]$ for processing the countdown $i $ line, the penultimate $i $ line missing number is $j $ of the scheme number.

Then there are:

$ $Dp [I][j] = \sum_{k=max (0,j-1)}^{m}dp[i-1][k]$$

Draw a drawing on your own to understand, here does not explain. After all, Gromah is too lazy ($ru \grave{o}$)

And then we thought about moving this picture out:

Then it's the number of path bars from top left to bottom right in this figure. (only left or right or down at a time)

Conversion, in fact, is to ask for this thing:

From $ (0,0) $ to $ (n*2+m+1,m+1) $, each time you can $x +1,y-1$ or $x +1,y+1$, and do not pass through $y =0$ and $y =m+1$ the number of path bars for both lines.

First, the complete works are ${n*2+m+1 \choose m+1}$,

Then we count the number of paths through the $y =0$, since through $y =0$ must pass through the $y =-1$, so we let the end and $y =m+2$ this line along $y =-1$,

You can then calculate the number of path bars (0,0) $ to the end of the rollover.

So it's not over yet. And those paths that go through $y =0$ and then through $y =m+1$ this straight line we're going to add back ...

It then flips the coordinate system along the $y =m+2$ (which should be $y =-m-4$) after flipping. Then the answer to the statistics ...

Until the scheme is $0$.

Calculate the number of path strips passing through the $y =m+1$ ...

I know my language ability and low, so the code is good ...

1#include <cmath>2#include <cstdio>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7typedefLong LongLL;8 #defineN 30000009 #defineMod 1000000007Ten  One intN, m, ans =0; A intFac[n +1], Inv[n +1]; -  -InlineintPowerintUintv) the { -     intres =1; -      for(; v; v >>=1) -     { +         if(V &1) res = (LL) res * U%Mod; -U = (LL) u * U%Mod; +     } A     returnRes; at } -  -InlinevoidPrepare () - { -fac[0] = inv[0] =1; -      for(inti =1; I <= N; i + +) inFac[i] = (LL) fac[i-1] * I%Mod; -Inv[n] = Power (fac[n), Mod-2); to      for(inti = N-1; I I--) +Inv[i] = (LL) Inv[i +1] * (i +1) %Mod; - } the  *InlineintCintUintv) $ {Panax Notoginseng     if(U <0|| V <0|| U < v)return 0; -     return(LL) fac[u] * Inv[v]% Mod * Inv[u-v]%Mod; the } +  AInlineintTintUintv) the { +     if(U < ABS (v))return 0; -     returnC (U, u-abs (v) >>1); $ } $  -InlineintINC (intUintv) - { the     returnU + V-(U + v >= Mod?) Mod:0); - }Wuyi  the intMain () - { Wu #ifndef Online_judge -Freopen ("4005.in","R", stdin); AboutFreopen ("4005.out","W", stdout); $     #endif -      - Prepare (); -scanf"%d%d", &n, &m); A     intx = n *2+ M +1; +Ans = T (x, M +1); the      -      for(inty = m +1, y_1 =-1, y_2 = m +2; X >=abs (y);) $     { they =2* Y_1-y; theY_2 =2* Y_1-y_2; theAns = Inc (ans, Mod-T (x, y)); they =2* Y_2-y; -Y_1 =2* Y_2-y_1; inAns =Inc (ans, T (x, y)); the     } the      About      for(inty = m +1, y_1 = m +2, y_2 =-1; X >=abs (y);) the     { they =2* Y_1-y; theY_2 =2* Y_1-y_2; +Ans = Inc (ans, Mod-T (x, y)); -y =2* Y_2-y; theY_1 =2* Y_2-y_1;BayiAns =Inc (ans, T (x, y)); the     } the      -printf"%d\n", ans); -      the #ifndef Online_judge the fclose (stdin); the fclose (stdout); the     #endif -     return 0; the}

4005_gromah

Bzoj 4005 [jloi 2015] cheat me.