TC Srm 597 Div 1 T3

Test instructions

Give M,r,g,b,2m=r+b+g,m represents a 2*m area of the number of columns, RGB Code Red Green and blue lattice number, requirements: Each 2*2 lattice, three colors each has at least one lattice, and the adjacent lattice color is different, ask how many kinds of arrangement

First we find that we can divide a two-grid placement scheme into two groups, {RB,BG,GR},{BR,GB,RG}

A legitimate scheme must consist of only one group, so we can calculate a set of scenarios and then x2 it

So we assume that RB has X, BG has y, GR has z, then ans is equivalent to: the number of placement schemes that make BR,RG,GB separate from itself.

Let's assume x>y>z, first we're going to use y,z to fill the gap between X,

If y+z<x-1, it cannot be filled, returns 0

Otherwise, we have 4 kinds of programs, fill in the middle x-1, X on the left, X on the right, all x+1, we can calculate the x-1~x+1 respectively

So now there's NX (nx=x+1/x/x-1) empty, we're going to fill it with y,z,

1. First of all, we assume that the Y1 is filled with Y, the scheme is C (nx,y-1), filled with Z z1=x-y1 empty

2. More y2=y-y1, the scheme of Y2 is equivalent to dividing all y into y1 parts, C (y-1,y1-1)

For each of the multi-plug y, we fortress a Z to prevent y adjacent, then after the end of our z is z2=z-z1-y2, and then insert these z y1 sequence of two ends, the scheme is C (2*Y1,Z2)

At this point, the assignment ends

#include <cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<cstring>using namespaceStd;typedefLong Longll;intn=1000000;intmo=1000000007;intjc[1000011],fc[1000011];intans,tmp,ts,x,y,z,xl,yl,req,zl,dt,tj,m,r,g,b,j,i;intMiintXintz) {    intl; L=1;  while(z) {if(z%2==1) l= (LL) l*x%mo; X= (LL) x*x%mo; Z/=2; }    returnl;}intCintNintm) {    return(LL) jc[n]*fc[m]%mo*fc[n-m]%mo;}intMain () {jc[0]=1;  for(i=1; i<=n;i++) jc[i]= (LL) jc[i-1]*i%mo; Fc[n]=mi (jc[n],mo-2);  for(i=n-1; i>=0; i--) fc[i]= (LL) fc[i+1]* (i+1)%mo; scanf ("%d",&DT);  for(tj=1; tj<=dt;tj++) {scanf ("%d%d%d%d",&m,&r,&g,&B); X=m-b;y=m-r;z=m-G; if(x<0|| y<0|| z<0) {printf ("0\n"); Continue; } ans=0; if(y>×) swap (x, y); if(z>x) swap (X,Z); if(z>y) Swap (y,z); if(y+z<x-1) {printf ("0\n"); Continue; }         for(xl=x-1; xl<=x+1; xl++){            if(Y+Z&LT;XL)Continue; TMP=0;  for(j=1; j<=y;j++) {req=xl-J; if(req>z)Continue; YL=y-K; if(req+yl>z)Continue; TS=C (XL,J); TS= (ll) ts*c (y1, J-1)%mo; ZL=z-req-yl; if(zl>2*J)Continue; TS= (ll) ts*c (2*J,ZL)%mo; TMP= (tmp+ts)%mo; }            if(xl==x) tmp=tmp*2%mo; Ans= (ans+tmp)%mo; } ans= (ans*2)%mo; printf ("%d\n", ans); }}

TC Srm 597 Div 1 T3