Common Boardimport sys n,k=map(int,input().split()) ans=0 nn=1 on=1 if n==1: print(0) sys.exit() for i in range(100000000000000000000): if on<=k: nn+=on on+=on ans+=1 elif k<n: nn+=k on+=k ans+=1 if nn>=n: print(ans) sys.exit() i can not understand why optimization my cod if i got ac with you help i give you 1000 rubles Edited by author 17.10.2018 22:35 The reason why you got time limit is that you don't use optimal algorithm. Your algorithm can be even correct, but to pass you need downgrade your algorithm. Maybe you could find some formula that would describe the whole process. Python is not the case here. You would get the same time limit issue with C. I launched your script with N = 10000000 and K = 2 and it took 1.833s on my machine to get the correct answer when the time limit is 0.25s. Of course it will grow up nonlinearly with bigger N. So it would take ages with N = 10^9 (the edge value). $ time echo "10000000 2" | python 1131_bruteforce.py real 0m1.833s user 0m1.824s sys 0m0.007s if i got ac with you help i give you 1000 rubles Edited by author 17.10.2018 22:35 P.S. I don't need your money if you finally solve it :) Edited by author 13.02.2020 06:12 Edited by author 13.02.2020 06:12Don't make 100+ submissions like I did. Just check these tests to make sure you correctly understand the problem description. Test #1: 10 20 0 5 3 4 turn 1 100 2 100 3 100 4 100 5 100 6 1 calls 20 2 calls 20 3 calls 20 4 calls 10 5 checks dealing flop 80 80 80 80 80 answer: 6 4 checks 5 checks 1 checks 2 checks 3 checks dealing turn Test #2: 10 20 10 5 3 3 turn 1 11 2 12 3 100 4 100 5 100 7 1 calls 1 2 calls 2 3 raises 10 to 30 4 folds 5 calls 10 dealing flop 5 checks 0 0 50 80 45 answer: 4 3 bets 10 5 calls 10 dealing turn 5 bets 5 Test #3 (this is what 10th test about): 10 20 0 5 3 4 flop 1 200 2 100 3 100 4 200 5 100 3 1 calls 20 2 calls 20 3 raises 20 to 100 100 80 0 100 80 answer: 5 4 calls 90 5 folds 1 calls 80 2 folds dealing flop Test #4: (this is what 32th test about - pay attention to the order of players' turns) 10 20 0 2 2 1 river 1 1000 2 1000 10 2 calls 10 1 checks dealing flop 1 bets 20 2 raises 30 to 50 1 calls 30 dealing turn 1 checks 2 checks dealing river 900 830 answer: 2 1 bets 30 2 raises 70 to 100 Test #5: (This is what 33th test about, 0 - is the wrong answer!!!!) 10 20 0 3 3 3 flop 1 100 2 100 3 100 0 80 80 80 answer: 6 3 calls 20 1 calls 10 2 checks dealing flop 1 checks 2 checks Good luck!!! Edited by author 12.06.2016 14:04 Edited by author 13.06.2016 20:51 Correct me if I'm wrong, but isn't test #2 invalid because it has one more possible answer? 5 3 checks dealing turn 5 checks 3 bets 10 5 raises 5 to 15 This answer isn't possible because of this: "If that is the first time when module takes turn then it should restore all players’ turns with all data it has. Otherwise module also has the information about all players’ turns before the previous bot turn". It means that you must have not more than one bot's turn in the answer. At least if M > 0. "It means that you must have not more than one bot's turn in the answer. At least if M > 0." Is there situation when M=0? Yes, first sample for example. But looks like there's always not more than one bot's turn. Here's some modification of my 5th test: 10 20 0 3 3 3 river 1 100 2 100 3 100 0 80 80 80 It seems to be correct, and the answer has > 1 bot's turn: 14 3 calls 20 1 calls 10 2 checks dealing flop 1 checks 2 checks 3 checks dealing turn 1 checks 2 checks 3 checks dealing river 1 checks 2 checks I checked and there are no such tests. "If that is the first time when module takes turn then it should restore all players’ turns with all data it has." - probably this means it's the first bot turn, otherwise the module would be called before. "Yes, first sample for example. But looks like there's always not more than one bot's turn." I mean M is number of all turns, not only bot's. Also I'd like to ask another question: in all tests here bot's turn occupies first line of output, so is there any test where bot's turn is not in first line, or it contradicts to the condition? Yes. See 1st sample test. And what about this test: 10 20 2 5 3 4 flop PhilIvey 500 TomDwan 500 ViktorBlom 500 Grobot 500 Ziigmund 500 6 PhilIvey folds TomDwan calls 20 ViktorBlom raises 40 to 60 Grobot calls 50 Ziigmund folds TomDwan folds 498 478 338 438 478 ans1: 1 dealing flop ans2: 3 dealing flop Grobot checks VictorBlom bets 100 Which answer is right? Thanks in advance. Edited by author 16.04.2017 17:27 for test #2 answer, wouldn't: 2 3 bets 10 5 raises 5 to 15 be correct too? and for test #3 raises shouldn't it be 3 raises 80 to 100 Thx in advance EDIT: Sorry, i misundestood Edited by author 02.12.2019 06:28 Edited by author 02.12.2019 06:28 Edited by author 21.02.2020 17:41 Easy concept, just to think for a while. AC in one go! just use bfs as a modified version of dijkstra and you will be done K= 11 S= 125 OTVET= 511435086865 K= 11 S= 126 OTVET= 620074954696 K= 11 S= 127 OTVET= 620074954696 K= 11 S= 128 OTVET= 747880479697 K= 11 S= 129 OTVET= 749351922670 K= 11 S= 130 OTVET= 900828406180 K= 11 S= 131 OTVET= 900828406180 K= 11 S= 132 OTVET= 1081759187586 K= 11 S= 133 OTVET= 1081759231344 K= 11 S= 134 OTVET= 1292739780552 K= 11 S= 135 OTVET= 1295226349065 I have got WA 23, but I know that my algo will have TLE... What is the right algo??? Accepted!!!! BFS - very good thing!!!! #include <iostream> #include <algorithm> #include <string> unsigned counter = 0; void printPalindroms(unsigned short *splitIndex, unsigned short i, const std::string& str){ if(splitIndex[i] == 0){ std::cout<<str.substr(0, i+1)<<" "; }else{ printPalindroms(splitIndex, splitIndex[i]-1, str); std::cout<<str.substr(splitIndex[i], i + 1 - splitIndex[i])<<" "; } } int main() { std::string strr; std::cin>>strr; const char* str = strr.data(); unsigned length = strr.size(); unsigned arrSize = (length*length + length)/2; bool *pArrData = new bool[arrSize]; bool **pArr = new bool*[length]; unsigned short *splitWeights = new unsigned short [length]; unsigned short *splitIndex = new unsigned short[length]; std::fill(pArrData, pArrData + arrSize, false); std::fill(splitWeights, splitWeights + length, length +2); std::fill(splitIndex, splitIndex + length, 0);
unsigned offset = 0; for(unsigned i = 0; i<length; ++i){ pArr[i] = pArrData + offset; offset += length - i; } // init the first row std::fill(pArr[0], pArr[0] + length, true); //init the second row for(unsigned i = 0; i<length - 1; ++i){ if(str[i] == str[i+1]){ pArr[1][i] = true; } } for(unsigned i = 2; i < length; ++i){ for(unsigned j = 0; j< length - i; ++j){ if(pArr[i-2][j+1] && str[j] == str[j+i]){ pArr[i][j] = true; } } } for(unsigned i = 0; i < length; ++i){ if(pArr[i][0] == true){ splitWeights[i] = 0; splitIndex[i] = 0; }else{ for(unsigned j = 1; j <= i; ++j){ if(pArr[i-j][j]){ unsigned short temp = splitWeights[j-1] + 1; if(temp < splitWeights[i]){ splitWeights[i] = temp; splitIndex[i] = j; } } } } } std::cout<<splitWeights[length -1] + 1<<std::endl; printPalindroms(splitIndex, length - 1, strr); delete[] splitIndex; delete[] splitWeights; delete[] pArr; delete[] pArrData;
return 0; } use cout.precision(0) and cout << sqrt(ans) * 1000.0 << endl; Edited by author 09.02.2020 20:13 How to solve this problem in o(n^2)? Let's start from the beginning: Let R(n) will be the function that returns count of unique staircases. Now let's introduce function G(n,j) that returns count of unique staircases each first step of which has more than j bricks. Then R(n) = G(n,0) - 1 . We have to subtract one because it is the case when all bricks are spent on the first step. G(n,j) = (n>j ? 1 : 0) + (G(n-j-1, j+1) + G(n-j-2, j+2) + ... +G(0,n)) G(n,j-1) - G(n,j) = (n == j ? 1 : 0) + G(n-j, j) => G(n,j) = G(n,j-1) - G(n-j,j) - (n == j ? 1 : 0) We know that in the case when n<=j G(n,j) = 0, so we can solve upper equation only for cases when j<n, in such cases upper formula will transform to G(n,j) = G(n,j-1) - G(n-j,j). But we still have to solve the cases when j == 0 and i > j as G(n, 0) = 1 + G(n-1, 1) + G(n-2, 2) + ... + G(0,n) //Alloc and init matrix g with zeroes //Calculat g matrix for(i = 2; i<=N; ++i){ g[i][0] = 1; for(j=1; j<i;++j){ g[i][0] += g[i-j, j]; } for(j=1; j<i;++j){ g[i][j] = g[i][j-1] - g[i-j,j]; } } cout<<g[N][0]-1; Edited by author 09.02.2020 17:14 I have WA 7 and I found the problem on test '20 30 13456'. It's a nice case to fix program but still it doesn't fix WA7. Edited by author 09.02.2020 15:46 Edited by author 26.03.2009 23:03 Edited by author 26.03.2009 23:03 There is full solution!!! Try thinking yourself or using hints before watching it and use visual c++ 2017 . . . . . . . . . . . . . . . . . . . //#include <bits/stdc++.h> #include <iostream> #include <vector> #include <algorithm> #define int long long #define pb push_back using namespace std; const int N = 50000; pair<int, int> dp[N], right_end[N]; vector<pair<int, int> > v; vector<int> ans; unsigned main() { for(int i = 0; i < N; i++) { right_end[i].first = -50001; right_end[i].second = -50001; } int m; cin >> m; int l, r; int k = 0; while(true) { cin >> l >> r; if(l > m || r < 0) continue; if(l == 0 && r == 0) break;
v.pb({l, r});
if(r > right_end[l].first) { right_end[l].first = r; right_end[l].second = k; } k++; }
int i = 0; for(auto it: v) { if(it.first <= 0 && it.second > 0) { if(it.second > dp[0].first) { dp[0].first = it.second; dp[0].second = i; } } i++; } for(int i = 1; i <= m; i++) { if(dp[i - 1].first >= right_end[i].first) { dp[i].first = dp[i - 1].first; dp[i].second = dp[i - 1].second; } else { dp[i].first = right_end[i].first; dp[i].second = right_end[i].second; } } for(int i = 0; i <= m; i++) { if(dp[i].first < i) { cout << "No solution" << endl; return 0; } } int pos = 0; while(pos < m) { ans.pb(dp[pos].second); if(dp[pos].first == pos) { cout << "No solution" << endl; return 0; } pos = dp[pos].first; } cout << ans.size() << endl; for(auto it: ans) { cout << v[(int)it].first << " " << v[(int)it].second << endl; } } Edited by author 09.02.2020 01:50 #include<iostream> using namespace std; float sqrt(long long2 n,int p) { int start = 0; int end = n; float ans; int mid; while(start <= end) { mid = (start+end)/2; if(mid*mid < n) start = mid + 1; else if(mid*mid > n) end = mid - 1; else { ans = mid; break; } } float increment = 0.1; for(int i = 0; i < p;i++) { while(ans * ans <= n) { ans += increment; } ans -= increment; increment /= 10; } return ans; } int main() { long long int n; while(cin >> n) { cout << sqrt(n,4) << endl; } } 1) Please read task, especially "from the last one till the first" carefully. Look at the example. 2) Your sqrt function lies. https://ideone.com/wcJVz0L=p^(q-1),p and q prima numbers. Please, help with this test. I am always getting WA on it. I did 2 assumptions: 1. If after any iteration the radius becomes irrational - then the answer will be irrational too. 2. The numerator and denominator of resulting fraction both don't exceed 2^64. Are these assumptions wrong? If so, please, give me some useful tests. The one interesting test I found is: 1 1 30 34785346 3131185849/1170753394542713625 But it doesn't help me much... Edited by author 04.02.2011 02:56 The 1st assumption is right. The 2nd one seems to be wrong. I can't say exactly if the result always fits in int64, but temporary (or itermediate) calculations exceed 2^64 for sure. Answer to your test is wrong. In this test the denominators of intermediate fractions (before reduction) exceed 2^64. My AC program answer: 1/6265197409 Oh, thanks, now I understood why it gets WA. Now my program gives correct answer to this test (I optimized it a little bit), but I found several tests with the same problem. It is interesting for me to solve it without BigInteger, I guess, there exists such solution for this problem. After a lot of tries I gave up :( AC now, but with big numbers. Edited by author 07.02.2011 23:02 Что бы не было переполнения вы могли сделать следующие : Представить r1 и r2 таким образом r1 = A / B ^ 2 r2 = A / C ^ 2 Где А = r1 * r2 / gcd(r1, r2); Отсюда найдете B и C. Теперь в чем прелесть такого представления: Вы наверное вывели формулу r3 = (r1 * r2 ) / (r1 + r2 + 2 * sqrt(r1, r2) ); Так вот если будете подставлять то получите r3 = A / ( B + C) ^ 2 т.е числитель не меняется а вот знаменатель будет меняться но не быстро ! I got Wa22(ac 1-21) under positions: r1*r2!=k*k-"irrational";(BUT IF i>min and i<MAX ONLY!) else 1/sqrt(r)=a/sqrt(r1)+b/sqrt(r2); a,b- recursion f(n,2i)=f(n-1,i),f(n,2i+1)=f(n-1,i)+f(n-1,i+1) All- __int64 AC now: a,b<=30- by induction. int- is enought, nor __int64, nor BigInteger Edited by author 27.02.2011 12:18 My mistake was: __int64 needed. a+b<=2^30 I have WA in the 9th test, please tell me this test. We have 4 case of perfect position of birds. I found the amount of flips for all cases. I used min_element() from algorithm.h to find the result, but it gave me WA 9. Then i replaced min_element() and found minimal by myself and my code finally accepted. Hope it will help you, good luck! Test case #6: 9 40 10 10 10 10 10 10 10 10 10 10 0 5 6 1 0 0 0 0 0 0 12 20 Answer: 158882255 Test case #12: 10 20 52 4 2 3 6 7 6 3 6 5 6 6 0 0 0 0 1 0 0 0 0 1 10 50 Answer: 978217740 PyCharm calculates n=2000 in 0.55 sec while here I got exceed time limit with 2.020 sec. I don't get it. Can't find the error, because it gives the same answers as those stupid precalc solutions: #include <bits/stdc++.h> using namespace std; using ll = long long; int soma[40] = {0}; int main() { ios::sync_with_stdio(0); cin.tie(0); int n, i, lim; ll ans = 0; cin >> n; lim = pow(10, n/2); for(i=0; i<lim; ++i){ soma[i%10 + (i/10)%10 + (i/100)%10 + (i/1000)%10]++; } for(i=0; i<=n*9; ++i){ ans += soma[i]*soma[i]; } cout << ans << '\n'; return 0; } for(i=0; i<=n*9; ++i){ ans += soma[i]*soma[i]; } i is out of soma range. #include<bits/stdc++.h> using namespace std; int n,m,q,a[100003],b[100003],c[100003]; vector<pair<int,int> >v; vector<int>g[100003]; bool vis[100003]; int qq[100003]; map<int,int>mp; void dfs(int x,int cc){ vis[x]=1; for(int i=0;i<g[x].size();i++) if(!vis[g[x][i]]) dfs(g[x][i],cc); qq[x]=cc; } int main(){ cin>>n>>m; for(int i=0;i<m;i++) cin>>a[i]>>b[i], a[i]--, b[i]--; cin>>q; for(int i=0;i<q;i++)cin>>c[i],c[i]--,v.push_back(make_pair(a[c[i]],b[c[i]])),a[c[i]]=b[c[i]]=-1; reverse(c,c+n);reverse(v.begin(),v.end()); for(int i=0;i<m;i++) if(a[i]!=-1) g[a[i]].push_back(b[i]), g[b[i]].push_back(a[i]); int cnt=0; for(int i=0;i<n;i++) if(!vis[i]) mp[cnt]=cnt, dfs(i,cnt++); vector<int>ans; for(int i=0;i<q;i++){ ans.push_back(cnt); if(mp[qq[v[i].first]]!=mp[qq[v[i].second]]) mp[qq[v[i].first]]=mp[qq[v[i].second]], qq[v[i].first]=qq[v[i].second], cnt--; } reverse(ans.begin(),ans.end()); for(int i=0;i<ans.size();i++) cout<<ans[i]<<' '; } are there any test? 5 8 2 3 1 2 2 5 1 4 1 5 3 4 3 4 4 4 8 7 6 8 2 3 4 5 1 JavaEntryPointWrapper.java:10: error: cannot access Oppa Oppa.main(args); ^ bad source file: .\Oppa.java file does not contain class Oppa Please remove or make sure it appears in the correct subdirectory of the sourcepath. 1 error Выходит такая ошибка, почему? У меня аналогичная ошибка. В том, что файл содержит один публичный класс, совпадающий с именем файла уверен на 100%. Причем проверил на файлах, которые отправлял ранее и удачно решенные, тоже самое. Даже решение из руководства не проходит... |
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