Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] E 三分+连续子序列的和的绝对值的最大值
2016-07-14 20:02
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E. Weakness and Poorness
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examples
Input
Output
Input
Output
Input
Output
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.
题意: 给你一段序列a1, a2, ..., an
a1 - x, a2 - x, ..., an - x. 对于每一个x 都有ans=连续子序列的和的绝对值的最大值
输出min(ans)
题解: 贪心求出连续子序列的和的绝对值的最大值 o(n)处理
三分x (x为实数存在负数) 求min(ans)
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examples
Input
3 1 2 3
Output
1.000000000000000
Input
4 1 2 3 4
Output
2.000000000000000
Input
10 1 10 2 9 3 8 4 7 5 6
Output
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.
题意: 给你一段序列a1, a2, ..., an
a1 - x, a2 - x, ..., an - x. 对于每一个x 都有ans=连续子序列的和的绝对值的最大值
输出min(ans)
题解: 贪心求出连续子序列的和的绝对值的最大值 o(n)处理
三分x (x为实数存在负数) 求min(ans)
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; int n; double a[200005]; double fun(double x) { double sum1=0.0,sum2=0.0; double max1=0.0,min1=100005.0; for(int i=1;i<=n;i++) { if((sum1+a[i]-x)<0) sum1=0; else sum1=sum1+a[i]-x; if((sum2+a[i]-x)>0) sum2=0; else sum2=sum2+a[i]-x; max1=max(max1,sum1); min1=min(min1,sum2); } return max(abs(max1),abs(min1)); } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lf",&a[i]); double l=-1e9,r=1e9,m1=0.0,m2=0.0; for(int i=0;i<600;i++) { m1=l+(r-l)/3.0; m2=r-(r-l)/3.0; if(fun(m1)<fun(m2)) r=m2; else l=m1; } printf("%f\n",fun(l)); return 0; }
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