UVa 12325 Treasure Chest

https://vjudge.net/problem/UVA-12325

Test instructions: There is a box with a volume of N and two infinite treasures. The volume of the Treasure 1 is S1, the value of V1 ' Treasure 2 of the volume of S2, the value of V2. Calculate the maximum amount of valuable treasures to be loaded.

The idea: The topic is very clear is the violent enumeration, but if does not simplify the enumeration the words certainly will be timed out, if N/S1 is smaller, then enumerates the number of Treasures 1, if N/S2 is smaller, enumerates the number of Treasures 2. There is also a situation is S1 and S2 are very small, S2 treasure 1 and S1 Treasure 2 of the same volume, and the value of S2*V1 and S1*v2 respectively. If the former is larger, then the Treasure 2 will only take s1-1 (otherwise S1 a treasure 2 to S2 1), if the latter is larger, then the Treasure 1 will only take s2-1.

1#include <iostream>2 using namespacestd;3 4 Long LongN, S1,v1, S2,v2;5 Long LongMAXN;6 7 voidsolve1 ()8 {9      for(inti =0; I <= n/s1; i++)Ten     { One         Long LongNumber = (N-I*S1)/S2; A         Long Longsum = i*v1 + number*V2; -         if(Sum > MAXN) maxn=sum; -     } the } -  - voidsolve2 () - { +      for(inti =0; I <= n/s2; i++) -     { +         Long LongNumber = (N-I*S2)/S1; A         Long Longsum = i*v2 + number*V1; at         if(Sum > MAXN) maxn=sum; -     } - } -  - voidsolve3 () - { in     if(s2*v1<s1*V2) -     { to          for(inti =0; I < S2 && i<=n/s1; i++) +         { -             Long LongNumber = (N-I*S1)/S2; the             Long Longsum = i*v1 + number*V2; *             if(Sum > MAXN) MAXN =sum; $         }Panax Notoginseng     } -     Else the     { +          for(intI=0; I < s1&&i<=n/s2; i++) A         { the             Long LongNumber = (N-I*S2)/S1; +             Long Longsum = i*v2 + number*V1; -             if(Sum > MAXN) MAXN =sum; $         } $     } - } -  the intMain () - {Wuyi     intN, Kase =0; theCIN >>N; -      while(n--) Wu     { -CIN >> N >> S1 >> V1 >> S2 >>V2; AboutMAXN =0; $         if(s1< -&& s2< -)  - solve3 (); -         Else if(N/s1 < N/S2) solve1 (); -         Elsesolve2 (); Acout <<"Case #"<< ++kase <<": "<< MAXN <<Endl; +     } the     return 0; -}

UVa 12325 Treasure Chest