Yes, I have gone though this. Its not in detail. On Fri, Mar 6, 2015 at 7:38 AM, Carlos Guia <[email protected]> wrote:
> Have you read the editorial? > http://community.topcoder.com/tc?module=Static&d1=match_editorials&d2=tco03_online_rd_4 > It's not very detailed, but should give enough insight. > > Carlos Guía > > On Mon, Mar 2, 2015 at 11:08 PM, Rajesh Kumar <[email protected]> > wrote: > >> Hi, >> >> I am trying very hard to understand the logic/theory of the solution to >> the problem TopCoder "Jewelry - 2003 TCO Online Round 4". Here is the >> problem statement:- >> >> Problem Statement >> >> You have been given a list of jewelry items that must be split >> amongst two people: Frank and Bob. Frank likes very expensive jewelry. Bob >> doesn't care how expensive the jewelry is, as long as he gets a lot of >> jewelry. Based on these criteria you have devised the following policy: •1) >> Each piece of jewelry given to Frank must be valued greater than or equal >> to each piece of jewelry given to Bob. In other words, Frank's least >> expensive piece of jewelry must be valued greater than or equal to Bob's >> most expensive piece of jewelry. >> •2) The total value of the jewelry given to Frank must exactly equal the >> total value of the jewelry given to Bob. >> •3) There can be pieces of jewelry given to neither Bob nor Frank. >> •4) Frank and Bob must each get at least 1 piece of jewelry. >> Given the value of each piece, you will determine the number of different >> ways you can allocate the jewelry to Bob and Frank following the above >> policy. For example: values = {1,2,5,3,4,5} >> Valid allocations are: Bob Frank >> 1,2 3 >> 1,3 4 >> 1,4 5 (first 5) >> 1,4 5 (second 5) >> 2,3 5 (first 5) >> 2,3 5 (second 5) >> 5 (first 5) 5 (second 5) >> 5 (second 5) 5 (first 5) >> 1,2,3,4 5,5 >> Note that each '5' is a different piece of jewelry and needs to be >> accounted for separately. There are 9 legal ways of allocating the jewelry >> to Bob and Frank given the policy, so your method would return 9. >> >> >> Here is the C++ solution:- >> #include <algorithm> >> #include <cstdio> >> #include <cstdlib> >> #include <cctype> >> #include <cmath> >> #include <iostream> >> #include <sstream> >> #include <string> >> #include <vector> >> #include <utility> >> using namespace std; >> >> #define REP(i,n) for(int i=0;i<(n);++i) >> #define FOR(i,a,b) for(int i=(a);i<=(b);++i) >> #define FORD(i,a,b) for(int i=(a);i>=(b);--i) >> #define FOREACH(i,c) for(__typeof((c).begin()) >> i=(c).begin();i!=(c).end();++i) >> typedef long long LL; >> const int INF = 1000000000; >> typedef vector<int> VI; >> template<class T> inline int size(const T&c) { return c.size(); } >> >> char buf1[1000]; >> >> string i2s(int x) { >> sprintf(buf1,"%d",x); >> return buf1; >> } >> >> const int MAXN = 30; >> const int MAX = 30000; >> >> int n; >> VI v; >> >> LL B[MAXN+1][MAX+1]; // [n pocz][sum] >> LL F[MAXN+1][MAX+1]; >> >> LL nk[MAXN+1][MAXN+1]; >> >> void cnk() { >> nk[0][0]=1; >> FOR(k,1,MAXN) nk[0][k]=0; >> FOR(n,1,MAXN) { >> nk[n][0]=1; >> FOR(k,1,MAXN) nk[n][k] = nk[n-1][k-1]+nk[n-1][k]; >> } >> } >> >> void calc(LL T[MAXN+1][MAX+1]) { >> T[0][0] = 1; >> FOR(x,1,MAX) T[0][x]=0; >> FOR(ile,1,n) { >> int a = v[ile-1]; >> FOR(x,0,MAX) { >> T[ile][x] = T[ile-1][x]; >> if(x>=a) T[ile][x] +=T[ile-1][x-a]; >> } >> } >> } >> >> struct Jewelry { >> // MAIN >> long long howMany(vector <int> vv) { >> v = vv; >> n = size(v); >> cnk(); >> sort(v.begin(),v.end()); //,greater<int>()); >> calc(F); >> sort(v.begin(),v.end()); >> calc(B); >> LL res = 0; >> int done=0; >> while(done<n) { >> int p=done; >> while(p<n && v[p]==v[done]) ++p; >> >> int c=p-done; >> FOR(u,1,c) { >> int uu = u * v[done]; >> FOR(x,uu,MAX) { >> res += B[done][x-uu] * F[n-done-u][x] * nk[c][u]; >> } >> } >> done=p; >> } >> return res; >> } >> }; >> >> >> Can some genius DP programmer make me understand the logic of this >> solution?? >> >> Regards, >> Rajesh Kumar >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Google Code Jam" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> To post to this group, send email to [email protected]. >> To view this discussion on the web visit >> https://groups.google.com/d/msgid/google-code/CALBPbNH_J2qV7chRLABDb0NKe%3DQeyo_LfA7U3bcQJvw9g3j91A%40mail.gmail.com >> <https://groups.google.com/d/msgid/google-code/CALBPbNH_J2qV7chRLABDb0NKe%3DQeyo_LfA7U3bcQJvw9g3j91A%40mail.gmail.com?utm_medium=email&utm_source=footer> >> . >> For more options, visit https://groups.google.com/d/optout. >> > > -- > You received this message because you are subscribed to the Google Groups > "Google Code Jam" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msgid/google-code/CANxkOGY_4X86deeO7afPZFD59rcpiwfYd%2BYjYDLZBwS0pNb9Dw%40mail.gmail.com > <https://groups.google.com/d/msgid/google-code/CANxkOGY_4X86deeO7afPZFD59rcpiwfYd%2BYjYDLZBwS0pNb9Dw%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > For more options, visit https://groups.google.com/d/optout. > -- Rajesh Kumar -- You received this message because you are subscribed to the Google Groups "Google Code Jam" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/google-code/CALBPbNGcoRyyEr0Z737yYN90gJ_rQeqf7tzC2xUQ0D3%2B11sHGg%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
