Show all threads Hide all threads Show all messages Hide all messages | I didn't thoght it would be so easy | SIAL_alky | 1573. Alchemy | 29 Jul 2018 22:56 | 3 | just the rule of multiplication in combinatorics... just one problem - the sequence of colors, but many 'if's will save the World! Use set data structure Add each color from input to a set Check the set for red, blue and yellow | Can somebody give me the whole explanation to this problem. Thanks beforehand! | Sanatbek_Matlatipov | 1573. Alchemy | 24 Nov 2016 16:01 | 2 | Can somebody give me the whole explanation to this problem. Thanks beforehand! If you have two different sets A = {a1, a2, a3} and B = {b1, b2, b3, b4} then we have |A| * |B| = 3 * 4 = 12 ways to choose a pair (a[i], b[j]). i\j b[1] b[2] b[3] b[4] a[1] [1,1] [1,2] [1,3] [1,4] a[2] [2,1] [2,2] [2,3] [2,4] a[3] [3,1] [3,2] [3,3] [3,4] Modern language solution looks like that: map<string, int> cnt; for (auto color : {"Blue", "Red", "Yellow"}) cin >> cnt[color]; int colors; cin >> colors; int ways = 1; for (int i = 0; i < colors; i++) { string color; cin >> color; ways *= cnt[color]; } cout << ways; Edited by author 24.11.2016 16:04 Edited by author 24.11.2016 16:07 | for wrong #4 | Bahodir | IT of TUIT | | 1573. Alchemy | 7 Mar 2015 22:44 | 1 | if(s.equals("Blue")|| s.equals("blue")) if(s.equals("Red") || s.equals("red")) if(s.equals("Yellow") || s.equals("yellow")) | wa 5 | isolver | 1573. Alchemy | 22 Jul 2012 02:48 | 1 | wa 5 isolver 22 Jul 2012 02:48 | hint | amirani | 1573. Alchemy | 17 Jan 2012 15:35 | 1 | hint amirani 17 Jan 2012 15:35 | Can Smb Explain The Prob. Or The Sample Output | SoSimple | 1573. Alchemy | 17 Dec 2011 17:00 | 2 | Can Smb Explain The Prob. Or The Sample Output | State: Why? | orcchg | 1573. Alchemy | 22 Aug 2011 13:54 | 2 | I got AC, but while looking at your solutions, i have guessed following: Nobody paid attention, that here may be more, than 3 potions: Total number of types of any potion is equal to 15: B, R, Y; B + R = V (Violet) B + Y = G (Green) R + Y = O (Orange) Proceed: B + V = BV; R + V = RV; Y + V = YV; and etc: BO, RO, YO, BG, RG, YG - "expert" potions; Sandro can combine any type of potions and any number less or equal than 15; Folowing to the state of this problem, we should consider any recipes of interesting final potion, and these recipes could include any of K <= 15 reagents - other potions; We also should recognize, what type of advanced potions Sandro can prepare, using his start reagents; And add new occasions into the answer. Sry for English, but i think - w/o considering the above, actual solution isn't correct. Edited by author 22.08.2011 01:36 Edited by author 22.08.2011 01:36 WTF? In the recipe thtre could be only “Blue”, “Red”, or “Yellow”. No Blue-Violet or Yellow-Orange. So if u even make some "expert" poisons, you're not able to do needed poison. Just use initial poisons. | haven't read "Each word occurs at most once." in statement and coded long arythmetics for this easy task :D | waterlink | 1573. Alchemy | 14 Apr 2011 22:20 | 1 | | I didn't understand. | CHIDEMYAN SERGEY | 1573. Alchemy | 25 May 2010 21:54 | 8 | Please explain me this problem>Thank!!! Give me some tests,please,because I think I don't understand problem correctly,please.Thank in advance!!! Edited by author 16.02.2008 20:28 this problem seems hard at first, but it's very easy 2 4 5 3 Red Yellow Blue >40 3 4 5 3 Red Yellow Blue >60 3 4 5 3 Red Yellow >36 3 4 5 2 Red Yellow >20 3 4 5 2 Blue Thank you for tests!!!I have just last questions:can k be more than 3(k>3)and in 3-th & 5-th of your tests why there is only 2(in 3-th test) and 1(in 5-th test) strings,while k=3(in 3-th test) and k=2(in 5-th test). Thank in advance!!! Edited by author 17.02.2008 00:30 no, k <= 3 and I gave you some wrong tests) 3 4 5 3 Red Yellow >36 - can't be(k must be 2) 3 4 5 2 Blue Red >12 Thank you very much!!!I get AC!!! Edited by author 17.02.2008 19:33 Test is wrong (Alexander (201 - P TNU)). If input: 3 4 5 2 Red Yellow Output must be 20, not 36. Edited by author 25.05.2010 21:56 | Might and Magic 8 is the best! | Komarov Dmitry [Pskov SPI] | 1573. Alchemy | 12 Nov 2009 12:06 | 2 | I think that this task describes the alchemy system from epic game M&M 8 (it's not 6 or 7, because in previous parts player couldn't play lich, but in 8 he could if he completed quest for necromancer). I love this game! Thanks to author, he caused nostalgia... Tell me, if I made mistake truying to identify the game. :) wiki page Komarov Dmitry [Pskov SPI] 12 Nov 2009 12:06 | why i got wa6!! | Time-warren258 | 1573. Alchemy | 18 Aug 2009 16:08 | 3 | here my code: #include<iostream> using namespace std; int main() { int flag[3]; int b,r,y; cin>>b>>r>>y; int n,i,j,k; char str[3][10]; cin>>n; i=0; while(i<n) cin>>str[i++]; for(i=0;i<n;i++) { if(str[i][0]=='B'||str[i][0]=='b') flag[0]=1; if(str[i][0]=='R'||str[i][0]=='r') flag[1]=1; if(str[i][0]=='Y'||str[i][0]=='y') flag[2]=1; } int sum=1; if(flag[0]==1) sum*=b; if(flag[1]==1) sum*=r; if(flag[2]==1) sum*=y; cout<<sum<<endl; return 0; } can anybody help me??? You didn't initialized "flag" and it contains random values. thanks,after i change "flag[3]=0;",i got a AC! |
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