Codeforce Gym 100685E Epic Fail of a Genie (greedy)

Test instructions: Give a bunch of elements, to find a subset, so that the product of the largest subset, if there are multiple, should make the subset of the number of elements as small as possible.

Puzzle: Greedy, if there is a positive number of products greater than 1, then it is necessary to choose, note that negative numbers may also be more than 1 positive number, then the absolute value is greater than 1 negative 22 pairs, if there is still one, then in the judgment, that negative and smaller than its smallest negative product is greater than 1, if so select two After selecting all the numbers that may be larger than 1, the remaining number must be smaller than 1, then it is not selected.

If you cannot make up more than 1 of the number, then categorize the discussion.

Very easy to write wrong ...

#include <bits/stdc++.h>using namespacestd;Const intMAXN = 1e4+6;structnum{intA[MAXN]; intID[MAXN]; intR[MAXN]; intsz; Num () {sz=0;} voidPushDoubleXinti) {A[sz]= x* -; R[SZ]=sz; ID[SZ]=i; SZ++; }    int_upper_bound (intLintRintv) { while(l<s) {            intm = (l+r) >>1; if(A[R[M]]&GT;V) R =m; ElseL = m+1; }        returnL; }    int_lower_bound (intLintRintv) { while(l<R) {            intm = (l+r) >>1; if(A[R[M]]&GT;=V) R =m; ElseL = m+1; }        returnL; }   //int ID (int x) {return r[x];}    int operator[](intx) {        returnA[r[x]]; }}p,m;intANS[MAXN];#defineGetbound (L,r,v,which) \L= Which._lower_bound (0, which.sz,v); R= Which._upper_bound (0, which.sz,v);#defineADD (which,x) \Ans[sz++] =Which.id[which.r[x]];BOOLcmpintXintY) {returnP.A[X] <p.a[y];}BOOLCMP2 (intXintY) {returnM.A[X] <m.a[y];}intMain () {//freopen ("In.txt", "R", stdin);    intN scanf"%d",&N); BOOLZero =false; intPzero;  for(inti =1; I <= N; i++){        DoubleT; scanf ("%LF",&t); if(t>0) {P.push (t,i); }Else if(t<0) {M.push (-t,i); }Else if(!Zero) {Zero=true; Pzero=i; }} sort (P.R,P.R+p.sz,cmp); Sort (M.R,M.R+M.SZ,CMP2); intm1l,m1r,p1l,p1r; Getbound (M1L,M1R, -, M) Getbound (P1L,P1R, -, P)intSZ =0; intT;  for(t = p1r; t < P.sz; t++) Add (p,t)intOdd = (M.SZ-M1R) &1;  for(t = m1r+odd; t < M.sz; t++) Add (m,t)if(odd) {if(m1r>0&& m[m1r]/10000. *m[m1r-1] >1) {Add (M,M1R) Add (m,m1r-1)  }    }    if(!SZ) {        intPsz = p.sz, Msz =M.sz; if(psz) {if(msz>=2&& m[msz-1]*m[msz-2] > p[psz-1]* -) {Add (M,msz-1) Add (m,msz-2)            }Else{ADD (p,psz-1) }        }Else {            if(msz>=2) {Add (M,msz-1) Add (m,msz-2)            }Elseans[sz++] =Pzero; }} sort (Ans,ans+sz); printf ("%d\n%d", sz,ans[0]);  for(inti =1; I < sz; i++) printf ("%d", Ans[i]); Putchar ('\ n'); return 0;}

Codeforce Gym 100685E Epic Fail of a Genie (greedy)