标签：序列 correct err sid ret def you color names
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.
Input
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
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.
Example
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
Note
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.
The poorness:连续子序列的和的绝对值。
题解：ternary searc（三分查找）https://www.cnblogs.com/whywhy/p/4886641.html，很典型的三分题。x取极值时最小，否则都会增大，所以是一个凹函数。
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=2e5+5; 5 6 int n; 7 int a[maxn]; 8 9 double maxSum(double A[]){ 10 double ans=0,tem=0; 11 for(int i=1;i<=n;i++){ 12 tem+=A[i]; 13 if(tem<0) tem=0; 14 ans=max(ans,tem); 15 } 16 return ans; 17 } 18 19 double compute(double x){ 20 double b[maxn]; 21 for(int i=1;i<=n;i++) b[i]=a[i]-x; 22 double ans1=maxSum(b); 23 for(int i=1;i<=n;i++) b[i]=-b[i]; 24 double ans2=maxSum(b); 25 return max(ans1,ans2); 26 } 27 28 int main() 29 { scanf("%d",&n); 30 for(int i=1;i<=n;i++) scanf("%d",&a[i]); 31 double l=-1e4,r=1e4; 32 for(int i=1;i<=100;i++){ 33 double midl=(2*l+r)/3.0; 34 double midr=(2*r+l)/3.0; 35 if(compute(midl)<compute(midr)) r=midr; 36 else l=midl; 37 } 38 printf("%.7f\n",compute(r)); 39 }
Weakness and Poorness CodeForces - 578C
标签：序列 correct err sid ret def you color names
原文地址：http://www.cnblogs.com/zgglj-com/p/7875629.html
