Added Bangla Tutorial for LOJ-1012 Guilty Prince #415
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rebornplusplus
requested changes
Dec 16, 2022
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Hello @walleeva2018. Great efforts. The picture is really nice!
However, I feel that the tutorial consists mostly about implementation plans i.e. how you would write the code to solve it. It's better if you lose them and re-write as if you were summarize the solution without the codes and all; in plain sentences.
I see that there is already an existing tutorial in English (#108) and it's a good one. So what you could also do instead, is translate that one.
1012/bn.md
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| আমরা প্রথমে একটি nxm ক্যারেক্টার অ্যারে তে আমাদের পুরো গ্রিড স্টোর করবো | ||
| নিচের উদাহরণ এর মত একটি 2D অ্যারে তে আমরা ক্যারেক্টার গুলা নিব | ||
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| ইনপুট টা কে অ্যারে তে স্টোর করার কোড | ||
| ``` | ||
| cin>>m>>n; | ||
| string s; | ||
| for(int i=0; i<n; i++) | ||
| { | ||
| cin>>s; | ||
| for(int j=0; j<s.size(); j++) | ||
| { | ||
| A[i][j]=s[j]; | ||
| } | ||
| } | ||
| ``` | ||
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| যেহেতু আমাদের @ থেকে শুরু করতে হবে সুতরাং আমাদের কোন সেল এ @ আছে টা কিছু একটা তে স্টোর করে রাখলাম । উদাহরণ হিসাবে উপরের চিত্রে (১,১) সেল টা আমাদের যাত্রা শুরুর সেল | ||
| ``` | ||
| for(int i=0; i<n; i++) | ||
| { | ||
| cin>>s; | ||
| for(int j=0; j<s.size(); j++) | ||
| { | ||
| A[i][j]=s[j]; | ||
| if(A[i][j]=='@') | ||
| { | ||
| x=i; | ||
| y=j; | ||
| } | ||
| } | ||
| } | ||
| ``` | ||
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1012/bn.md
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| void dfs(int x,int y) | ||
| { | ||
| if(x==n || y==m ||x==-1 || y==-1) | ||
| return; | ||
| if(vis[x][y]==1) | ||
| return ; | ||
| if(A[x][y]=='#') | ||
| return ; | ||
| cnt++; | ||
| vis[x][y]=1; | ||
| dfs(x+1,y); | ||
| dfs(x,y+1); | ||
| dfs(x-1,y); | ||
| dfs(x,y-1); | ||
| } | ||
| ``` | ||
| ### কিছু সতর্কতা | ||
| প্রতি কেস এর পর vis এবং cnt ক্লিয়ার করতে হবে |
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Okay Sir , I am fixing it
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I tried to fix it Sir. Can you review it and tell me what to do next
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