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Suggested Solutions of Sharif Internet Contest 09 Aideen NasiriShargh (aideen.ir) 11/5/2009

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Problem A: AmaTeuR HaCKeRS!

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YES SIGNAL NO SIGNAL You ARe HaCKeD!You ARe SiBiLiZeD! Should I tell the rest?!

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Problem A: AmaTeuR HaCKeRS! Author: Aideen NasiriShargh Solution & TestData: Aideen NasiriShargh Task: Count number of HaCKeRiSH words in a line! 117 out of 135 teams submitted. 57 teams got AC. Using stringstream, one could get away from all end-line (CR-LF) problems. Although it was a Task-A problem, STL (in C++) could come to help.

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Problem A (cont’d) bool ishacker(const string &s) { if (!isupper(s[0])) return false; FRI(it,s) { if (it == s.begin()) continue; bool vowel = (string("aeiou").find(tolower(*it)) != string::npos); if (vowel ^ (bool)islower(*it)) return false; } return true; } #define FOR(i,x,n) for (__typeof(x) i = (x); i != (n); i++) #define FR(i,n) FOR(i,0,n) #define FRI(it,v) FOR(it,v.begin(),v.end())

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For each testcase { int n = 0, good = 0; getline(cin, s); stringstream ss(s); while (ss >> w) { n++; good += ishacker(w); } printf("%d out of %d.\n", good, n); }

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Problem B: Bahman’s Disapproval! Author: Aideen NasiriShargh Solution & TestData: Aideen NasiriShargh Task: Count number of prime numbers in a range of 100K numbers. 98 out of 135 teams submitted. 44 teams got AC. Important Points: ◦ Range was fixed [88’200K, 88’201K), each one’s prime-ity can be calculated in O(√n) but ◦ Number of Tests (1 ≤ T ≤ 100) could cause redundancy!

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Problem B (con’td) Solution: ◦ Calculate isPrime[x] for all x’s in the range once and store it later. Optional Improvement (you could get AC w/o this one) ◦ Store all prime numbers in a sorted-array and answer each query in O(lg K) where K is number of primes in the range (near 5% of 100K).

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Problem B (con’td) const int Base = 88100000; const int MAX_W = 199999; bool prime[MAX_W+3]; bool isprime(int x) { if (x == 2) return true; if (x <= 1 || x % 2 == 0) return false; for (int i = 3; i * i <= x; i += 2) if (x % i == 0) return false; return true; }

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Problem B (con’td) int main() { int testn; cin >> testn; FR(i,MAX_W+1) prime[i] = isprime(Base+i); FR(testi,testn) { int a, b; cin >> a >> b; int ans = 0; FOR(i,a,b+1) ans += prime[i-Base]; //isprime(i); double p = 100 * double(ans) / (b-a+1); printf("%0.0lf%\n", p); } return 0; }

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Problem C: Captivity Of Casuality! Author: SaeedReza Seddighin Solution & TestData: SaeedReza Seddighin Task: Given some “X because Y”, you should answer “Why Z?”. Causes are transitive. 40 out of 135 teams submitted. 25 teams got AC. Solution: ◦ Separate sentences by “because” to get X and Y. ◦ Add this clue as a directed edge from Y to X. ◦ For each question, follow the path from Z as long as you can.

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Problem C: Captivity Of Casuality! Idea: Use STL! ◦ map C; // The graph as a map ◦ Set S; // To store current X’s while (S.find(x) == S.end()) if (C.find(x) == C.end()) break; else { S.insert(x); x=Map[x];} if (Set.find(x) == Set.end()) cout<<"Because "<

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Problem D: Dr. B! Author: SaeedReza Seddighin Solution & TestData: SaeedReza Seddighin Task: Let Dr. B apply his fans! 4 out of 135 teams submitted. 2 teams got AC. Important Points: ◦ The main idea for solving this problem is a full search. ◦ we should test all possible combinations of classes that could be our result, and choose the one which has most difference with initial ones.

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Problem D: Dr. B! (cont’d) How? ◦ First we should generate all possible combinations ( which are C(m,n) ). ◦ Backtracking is an efficient way for generating but there are some other simpler ways that coder may use. ◦ After that we should check whether a combination a valid or not, validity means that Dr. B could make that combination using his rules. Now the problem is what a valid combination is. ◦ It is obvious that classes of a valid combination should be independent to each other. In the other hand no two classes should share a period of time. ◦ Second condition is that there should be a way to add/remove classes that changes initial combination of classes into it.

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Problem D: Dr. B! (cont’d) Condition 1) Classes of a valid combination should be independent to each other. In the other hand no two classes should share a period of time. ◦ Testing the first condition could be easilty done in o(n*n) time. Condition 2) There should be a way to add/remove classes that changes initial combination of classes into it. ◦ For testing the second condition we can consider all m! ways of inserting new classes. it is a tricky point that a way of inserting classes is valid (could be done using rules) iff after inserting i th class at least i classes of initial combination should have conflict with new ones.!

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Problem E: Ehem, Ehem, NBL! Author & Solution: SaeedReza Seddighin TestData: Aideen NasiriShargh Task: We know one person is either Lucky or Unlucky. A team is a group of 3 persons. A team is Winner iff all 3 persons in that team be Lucky. 17 out of 135 teams submitted. 11 teams got AC. Is it the famous NP-complete problem of 3-SAT ? ◦ No! 3Sat is (A OR B OR C) AND (X OR Y OR Z) ◦ But, here we have (A AND B AND C)!

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Problem E: Ehem, Ehem, NBL! Solution: ◦ Consider all winner teams, all members of those teams MUST BE lucky. (isn’t it?) Set their luck==true. ◦ Now, after the previous step, consider the rest (losers team). Is there any team among them, which has 3 members with who are luck==true? If so, it’s an inconsistency. Otherwise, you can call persons with unset luck, unlucky!

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Problem E (cont’d) How to find the state of the new team? ◦ Easy! Assume it is lucky, run the inconsistency test once again (now in n+1 teams). Any contradiction means the new team is loser. ◦ Otherwise, the new team can be winner!

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Problem E (cont’d) bool solve (int n) {// is first n teams consistent? set lucky; FR(i,n) if (a[i].gold) // winners FR(j,3) lucky.insert(a[i].mem[j]); FR(i,n) if (!a[i].gold) { // losers FR(j,3) if (!isin(lucky, a[i].mem[j])) goto hell; return false; // a loser with 3 not-goto-ed men hell: ; // check next loser team } return true; }

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Problem E (cont’d) if (!solve(n)) cout << "Inconsistent data!" << endl; else if (!solve(n+1)) cout << "No, they are natural-born- losers!" << endl; else cout << "Yes, they can win!" << endl;

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Problem F: Factorialz Again! TestData & Solution: Mahdi SafarNejad Task: Find number of trailing zeros (in right end) of n! in base K. 53 out of 135 teams submitted. 23 teams got AC. It’s about math again! Random tests causes some incorrect codes get AC. :-(

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Problem F: Factorialz Again! The Sieve of Eratosthenes is a simple algorithm for finding all prime numbers up to a specified integer. (from wikipedia)

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Problem F: Factorialz Again! Solution: ◦ For each prime number P, find maximum k that: n % (P^k) == 0 How do you do that? ◦ Answer is minimum of such k’s for all Ps. Incorrect solution: ◦ Do that only for largest available P, not all Ps. As you work on powers of 5 when base is 10.

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Problem G: Glamorous Polygons! Author, Testdata, Solutions: Sepideh MahAbadi Eased Solution: Aideen NasiriShargh Task: Given 2 Polygon, tell whether they are similar in shape or not? 9 out of 135 teams submitted. 2 teams got AC. N can be as large as 100’000. So O(N^2) algorithm does not work!

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Problem G: Glamorous Polygons! Solution: ◦ Two polygons are similar if: 1) Their angels be the same in order. (is it enough)? Imagine Rectangle and Square! 2) Length of their edges have the same ratio. (is it enough)? Image lozenge and Square! ◦ Both of the above conditions must be held. Why is it enough now?

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Problem G: Glamorous Polygons! Solution: ◦ Calculate pair for both of the polygons and store them in two arrays. Now you want to find other whether one can become like the other by shifting and multiplying the lengths in an arbitrary factor? ◦ How can you do that in O(N)? duplicate the first array and append the copy to it’s end. i.e. ABC => ABCABC Now find the second array (size N) in the first array (which is now 2N in size). Use that by KMP algorithm in O(N + 2N) = O(N).

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Problem H: Horrbly Fast-Calc! TestData & Solution: Mahdi SafarNejad Task: Find sum of all positive divisors of numbers in a range. 99 out of 135 teams submitted. 61 teams got AC. Good Idea: ◦ Find each primes’s impact in final result one by one. ◦ Slow codes could get AC as someone was generous!

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Problem I: Intergalactic souvenirs! Author: Aideen NasiriShargh TestData & Solution: Aideen NasiriShargh Task: Given some pre-comparisons of some objects; you are to find minimum and maximum of them (in weight) with the least number of balances. 4 out of 135 teams submitted. 9 teams got AC.

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Problem I: Intergalactic souvenirs! First of all, is it inconsistent? ◦ Have a directed graph, you want to find if it has a cycle or not in O(N+E). You need to have 3 colors and find double-reached gray vertex. Second, find minimum number of additional weightings. ◦ For each object we have CanBeMin[x] and CanBeMax[x] which are all initially true.

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Problem I: Intergalactic souvenirs! After given comparisons, objects are in 4 type: 1.A: CanBeMin and CanBeMax 2.B: CanBeMin and !CanBeMax 3.C: !CanBeMin and CanBeMax 4.D: !CanBeMin and !CanBeMax Drop D’s! ◦ Each comparison between X and Y can make !CanBeMin[X] and !CanBeMax[Y] (if X > Y) !CanBeMax[X] and !CanBeMin[Y] (if Y > X)

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Problem I: Intergalactic souvenirs! Each object in B or C need at least one comparison. Each object in group A needs one comparison and then one comes to B and one comes to C. So the answer is ◦ A + (B+A/2 -1) + (C+A/2 -1) ◦ If B=C=0 then answer is ceil(3N/2)

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Any Question?

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