Xjoi Online synchronization Training DAY1 T3

Idea: At the beginning to see this question when want DP, but found seemingly not. Because there is a prefix and suffix, and some suffixes will overwrite the current prefix, this does not satisfy the no-no effect of the AH!

But there is a very ingenious idea: if we know the maximum value of a[i, then the number of P and the number of Q are also determined. So the sequence length is also determined, set m for the sequence length.

And for each a[i] represents a fixed number of p and Q and lengths.

Therefore, the prefix length is greater than M/2, we can subtract it with the total p and the total Q, and convert it to a prefix suffix less than or equal to the M/2 length.

So we can design DP for F[I][J][K], representing a J p from left to right I, and a k p for the left I, so the position of f[(m+1)/2] is the final answer!

Note that a pre array is recorded to record where the optimal solution for this state is transferred from.

1#include <algorithm>2#include <cstdio>3#include <cmath>4#include <cstring>5#include <iostream>6 #definell Long Long7ll a[200005];8 Constll pw=9705276;9 Constll qw=12805858;Ten intc[200005][2],n,pre[205][205][205][2],w[205][205],f[205][205][205]; One intans[200005],CN; A intRead () { -     CharCh=getchar ();intt=0, f=1; -      while(ch<'0'||'9'<ch) {if(ch=='-') f=-1; ch=GetChar ();} the      while('0'<=ch&&ch<='9') {t=t*Ten+ch-'0'; ch=GetChar ();} -     returnt*F; - } - voidinit () { +n=read (); -      for(intI=0; i<n;i++){ +         DoubleX;SCANF ("%LF",&x); AA[i]= (LL) (x*100000+0.5); at     } - } - voidsolve () { -     intmxpos=0; -      for(intI=1; i<n;i++){ -         if(A[i]>a[mxpos]) mxpos=i; in     } -     inttotp=-1, totq=-1; to      for(intI=0;(ll) i*pw<=a[mxpos];i++) +      if(a[mxpos]-(LL) I*PW)%qw==0){ -totp=i; theTotq= (int) ((a[mxpos]-(LL) I*PW)/QW); *              Break; $     }Panax Notoginseng      for(intI=0; i<n;i++){ -         intp=-1, q=-1; the          for(intj=0;(ll) j*pw<=a[i];j++) +          if(a[i]-(LL) J*PW)%qw==0){ Ap=J; theQ= (int) ((a[i]-(LL) J*PW)/QW); +                  Break; -          } $          if(p!=-1&&p+q<=totp+TOTQ) { $c[cn][0]=p; -c[cn][1]=Q; -cn++; the          } -     }Wuyi     intm=totp+TOTQ; the      for(intI=0; i<cn;i++){ -         if(c[i][0]+c[i][1]<=m/2){ Wuw[c[i][0]+c[i][1]][c[i][1]]++; -}Else{ Aboutw[m-c[i][0]-c[i][1]][totq-c[i][1]]++; $         } -     } -f[0][0][0]=0; -      for(intI=1; i<=m/2; i++) A       for(intj=0; j<=i;j++) +        for(intk=0; k<=i;k++){ the             ints=-1; -              for(intp=0;p <=1;p + +){ $                  for(intq=0; q<=1; q++){ the                     if(j-p>=0&&j-p<i&&k-q>=0&&k-q<i&&f[i-1][j-p][k-q]>R) { thes=f[i-1][j-p][k-Q]; thepre[i][j][k][0]=p; thepre[i][j][k][1]=Q; -                     } in                 } the             } thef[i][j][k]=s+w[i][j]+ (k==j)?0: W[i][k]); About       } the     intansi=-1, ansj=-1, ansk=-1; the      for(intk=0; k<=m%2; k++) the       for(intI=0; i<=m/2; i++){ +         intj=totq-k-i; -         if(j>=0&&j<=m/2&& (ansi==-1|| f[m/2][i][j]>f[m/2][ANSI][ANSJ])) { theAnsi=i;Bayiansj=J; theansk=K; the         } -     } -     if(m%2) ans[m/2]=Ansk; the      for(inti=m/2;i>0; i--){ the         intp=pre[i][ansi][ansj][0]; the         intq=pre[i][ansi][ansj][1]; theans[i-1]=p; -ans[m-i]=Q; theansi-=p; theansj-=Q; the     }94      for(intI=0; i<m;i++) the      if(Ans[i]) printf ("Q"); the      Elseprintf"P"); the }98 intMain () { About init (); - solve ();101}

Xjoi Online synchronization Training DAY1 T3