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Microsoft OA question

Правка en1, от algoskipper, 2024-07-12 20:31:20

This question is asked un Microsoft online assessment which is present in various sites but none of them have provided correct solutions. One can think of bfs but it will give TLE as many steps are repeated a lot of time. So talking about dp optimization. 2D dp where dp[i][j] will give total number of ships when start with value j and total layers are i including current layer. Base case would be for every i dp[i][0]=i and for every j dp[1][j]=1. dp[i][j]= (summation dp[i-1][j] for every j from 0 to (j*j+1)%m-1 ) But its constantly giving wrong answer. And to make it worse nowhere correct answer is given nor failed test case is shown. i think i have something wrong in my idea can anyone clarify. below is my code.

ll l,m;cin>>l>>m;
    vector<vector<ll>>dp(l+2,vector<ll>(m+2,0));
    f(m){dp[1][i]=1;}

    vector<ll>lp(m+1);lp[0]=1;
    for(ll i=1;i<l;i++){lp[i]=lp[i-1]+1;}
for(ll i=0;i<l;i++){dp[i][0]=i;}
    for(ll j=2;j<=l;j++){

for(ll i=0;i<m;i++){
    if(i==0)continue;
    if((i*i+1)%m!=0)dp[j][i]=lp[((i*i+1)%m)-1];
    dp[j][i]++;

}
for(ll i=0;i<lp.size();i++){if(i==0)lp[i]=dp[j][i];else lp[i]=lp[i-1]+dp[j][i];}

    }
    cout<<dp[l][2]%m<<endl;

question link Question link

Теги microsoft, assessment, bfs, dp

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  Rev. Язык Кто Когда Δ Комментарий
en3 Английский algoskipper 2024-07-12 20:40:58 76
en2 Английский algoskipper 2024-07-12 20:36:24 17
en1 Английский algoskipper 2024-07-12 20:31:20 1399 Initial revision (published)