CF578C weakness and poorness three points

In fact, three points is a single-peak function of the most value of things, usage is more uniform. The problem is that the observation found that the value is a single-peak function, then the enumeration of T three points on the line.

The problem is dry:

Given an array of length n, the AI, a real number x, so that the sequence a1-x,a2-x,..., an-x is the least good degree. The degree of uncertainty is defined as the maximum value of the bad degree of all successive sub-sequences of a sequence. The bad degree is defined as the absolute value of the sum of all the positions of a sequence. Enter the first row N. The second row n number, which represents AI. Outputs a real number that is accurate to 6 decimal places, indicating the minimum degree of uncertainty. 

Code:

#include <iostream>#include<cstdio>#include<cmath>#include<ctime>#include<queue>#include<algorithm>#include<cstring>using namespacestd;#defineDuke (I,a,n) for (int i = a;i <= n;i++)#defineLV (i,a,n) for (int i = a;i >= n;i--)#defineClean (a) memset (A,0,sizeof (a))Const intINF =1<< -; typedefLong LongLl;typedefDoubledb;template<classT>voidRead (T &x) {    CharC; BOOLOP =0;  while(c = GetChar (), C <'0'|| C >'9')        if(c = ='-') op =1; X= C-'0';  while(c = GetChar (), C >='0'&& C <='9') x= x *Ten+ C-'0'; if(OP) x =-x;} Template<classT>voidwrite (T x) {if(X <0) Putchar ('-'), x =-x; if(x >=Ten) Write (X/Ten); Putchar ('0'+ x%Ten);}Const intMAXN = 2e5 +5;ConstDB EPS = 1e-8;intARR[MAXN];d b arr1[maxn];intn;db Calc (db arr[]) {db T=0; DB Res= arr[0]; Duke (I,0N1) {T= Arr[i] +T; Res=Max (res,t); if(T <EPS) T=0; }    returnRes;} DB Solve (db t) {Duke (I,0N1) {Arr1[i]= Arr[i]-T; } DB x1=Calc (arr1); Duke (I,0N1) Arr1[i]= -Arr1[i]; DB X2=Calc (arr1); DB x=Max (Fabs (x1), fabs (x2)); returnx;}intMain () {read (n); intMaxnum =-10001, Minnum =10001; Duke (I,0N1) {read (arr[i]); Maxnum=Max (maxnum,arr[i]); Minnum=min (minnum,arr[i]); }    intdcnt = -; DB L= Minnum, R =Maxnum;  while(dcnt--) {db M1= L + (r-l)/3; db M2= R-(r-l)/3; DB X1=solve (M1); DB X2=solve (m2); if(X1 >x2) {L=M1; }        Else{R=m2; }} DB Res=solve (L); printf ("%.8lf\n", RES); return 0;}

CF578C weakness and poorness three points