[Discrete logarithm] ultraviolet A 11916 Emoogle Grid

[Discrete logarithm] ultraviolet A 11916 Emoogle Grid

Question:

Apply K colors to an M * N matrix.

There are B cells that cannot be colored, and each cell cannot be the same color as the cell above.

If N and total result R are known, find the row M with the minimum satisfaction.

All data has a modulo value of 100,000,007.

Ideas:

Because the X coordinates of the input vertex that cannot be colored must be within M. So record maxX.

Divide the entire matrix into three parts.

The first part is maxX * N. The answer in this part can be used to determine whether it is satisfied.

The second part is (maxX + 1) * N. This part has one more row. In order to remove the influence of some grids that cannot be entered.

The third part is M * N, followed by (M-maxX-1) line each lattice filled in color is only K-1.

At this time set the previous answer to ans, set M-maxX-1 = T

Then ans * (K-1) ^ (T * N) = R (mod 100000007)

At this time, we need to request an X ^ X = B (mod M)

An enumeration method is required.

Code:

 

#include"cstdlib"#include"cstdio"#include"cstring"#include"cmath"#include"queue"#include"algorithm"#include"iostream"#include"map"#define ll long longusing namespace std;int mark[502][502];map
 
  id;ll mod=100000007;int n,k,b,idcnt,maxx,maxcnt;ll r;ll power(ll a,ll b){    ll ans=1;    while(b)    {        if(b&1) ans=(ans*a)%mod;        a=(a*a)%mod;        b/=2;    }    return ans%mod;}ll inv(ll x){    return power(x,mod-2);}ll log_mod(ll a,ll b){    ll m,v,e=1,i;    m=(ll)sqrt(mod+0.5);    v=inv(power(a,m));    map
  
   x;    x.clear();    x[1]=0;    for(i=1;i
   
    =mod) e%=mod;        if(x[e]==0) x[e]=i;    }    for(i=0;i
    
     =mod) b%=mod;    }    return -1;}int solve(){    //1~maxx    ll ans=1;    if(maxx!=0)    {        for(int i=1;i<=idcnt;i++)        {            int cur=0;            mark[i][++mark[i][0]]=maxx+1;            for(int j=1;j<=mark[i][0];j++)            {                int sum=mark[i][j]-cur-1;                cur=mark[i][j];                if(sum==0) continue;                ans*=k;                if(ans>=mod) ans%=mod;                ans*=power(k-1,sum-1);                if(ans>=mod) ans%=mod;            }        }        ans*=power(k,n-idcnt);        if(ans>=mod) ans%=mod;        ans*=power(k-1,(maxx-1LL)*(n-idcnt));        if(ans>=mod) ans%=mod;        if(ans==r) return maxx;    }    //maxx+1    ans*=power(k,maxcnt);    if(ans>=mod) ans%=mod;    ans*=power(k-1,n-maxcnt);    if(ans>=mod) ans%=mod;    if(ans==r) return maxx+1;    ll tep=inv(ans);    ll A,B;    B=(tep*r)%mod;    A=power(k-1,n);    return log_mod(A,B)+maxx+1;}int main(){    int t,cas=1;    cin>>t;    while(t--)    {        scanf("%d%d%d%lld",&n,&k,&b,&r);        memset(mark,0,sizeof(mark));        id.clear();        idcnt=0;        maxx=0;        maxcnt=n;        for(int i=0;i
     
      maxx)            {                maxx=x;                maxcnt=1;            }            else if(x==maxx) maxcnt++;        }        printf("Case %d: %d\n",cas++,solve());    }    return 0;}