Codeforces Round #594 (Div. 1A) lvan the Fool and Probability Theory

The meaning of problems, only black and white of a matrix

The picture shows the probability of satisfying said each square up and down four directions only one color squares and squares the same color map of the current

answer

By meter

-0	2*1	2*2	2*3	2*5	2*8	2*13
-1	2*2	2*3	2*4	2*6	2*9	2*14
-2	2*3	2*4	2*5	2*7	2*10	2*15
-4	2*5	2*6	2*7	2*9	2*11
-7	2*8	2*9	2*10	2*12	2*15
-12	2*13	2*14	2*15	2*17

We found each row meet $ f [i] = f [i-1] + f [i-2] -c $ form

$ C $ conform $ abs (c [i]) = abs (c [i-1]) + abs (c [i 2-]) + 1 $ the form

The first two columns in line with $ f [i] = f [i-1] + f [i-2] -c $ form

Recurrence

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define A 1111111
const ll mod=1e9+7;
ll n,m,c;
ll cha[A],thefirst[A],thesecond[A],ans[A];
int main () {
    scanf("%lld%lld",&n,&m);
    father [ 1 ] = 0 ; father [ 2 ] = 1 ;
    thefirst[1]=1;thefirst[2]=2;
    thesecond[1]=2;thesecond[2]=3;//最后统一
    for(ll i=3;i<=100000;i++)
        cha[i]=(cha[i-1]+cha[i-2]+1)%mod;
    for(ll i=3;i<=100000;i++)
        thefirst[i]=(thefirst[i-1]+thefirst[i-2])%mod;
    for(ll i=3;i<=100000;i++)
        thesecond[i]=(thesecond[i-1]+thesecond[i-2]-1)%mod;
    ans[1]=thefirst[n]%mod;ans[2]=thesecond[n]%mod;
    c=cha[n];
//  printf("1=%lld 2=%lld c=%lld\n",ans[1],ans[2],c);
    for(ll i=3;i<=100000;i++)
        ans [i] = (ans [i- 1 ] + ans [i- 2 ] c)% v;
    printf("%lld\n",ans[m]*2%mod);
}

View Code