Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] E. Weakness and Poorness 三分
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.
题意:找出一个x,使得a[i]-x的这段连续最大子序列和绝对值最小;
思路:三分,check的时候来回扫一遍
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=2e5+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; double a[N],v[N]; int n; double equ(double x) { double ans=0; double sum=0; for(int i=1;i<=n;i++) { sum+=(a[i]-x); if(sum<0) sum=0; ans=max(ans,sum); } sum=0; for(int i=1;i<=n;i++) { sum-=(a[i]-x); if(sum<0) sum=0; ans=max(ans,sum); } return ans; } double ternarySearch(double l,double r) { for(int i=1;i<=100;i++) { double lll=(2*l+r)/3; double rr=(l+2*r)/3; double ans1=equ(lll); double ans2=equ(rr); if(ans1>ans2) l=lll; else r=rr; } return l; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lf",&a[i]); double ans=ternarySearch(-20000,20000); printf("%f\n",equ(ans)); return 0; }