You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
3 1 2 3
1.000000000000000
4 1 2 3 4
2.000000000000000
10 1 10 2 9 3 8 4 7 5 6
4.500000000000000
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
三分查找:如果遇到凸性或凹形函数时,可以用三分查找求那个凸点或凹点。
#include<bits/stdc++.h> using namespace std; #define rep(i,a,n) for(int i=a;i<n;++i) #define per(i,a,n) for(int i=n-1;i>=a;--i) #define mem(a,t) memset(a,t,sizeof(a)) #define pb push_back #define mp make_pair #define sz(a) (int)a.size() #define fi first #define se second typedef long long LL; #define N 200005 int a[N]; int n; const double eps=1e-12; double fun(double x) { double tmp=0,ans=0,tmin=0,tmax=0; rep(i,0,n){ tmp+=(a[i]-x); ans=max(ans,fabs(tmp-tmin)); ans=max(ans,fabs(tmp-tmax)); tmin=min(tmin,tmp); tmax=max(tmax,tmp); } return ans; } int main() { //freopen("in.txt","r",stdin); double l,r; scanf("%d",&n); l=1e7; r=-1e7; rep(i,0,n){ scanf("%d",&a[i]); l=min(l,1.*a[i]); r=max(r,1.*a[i]); } double f1,f2,mid1,mid2; rep(i,0,500){ mid1=l+(r-l)/3; mid2=r-(r-l)/3; f1=fun(mid1); f2=fun(mid2); if(f1<f2) r=mid2; else l=mid1; } cout.precision(10); cout<<fixed<<fun(l)<<endl; return 0; }