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 ： 寻找一个X ，让数组中的全部元素都减去 x ,找区间和最大的情况，所求答案是所有可能情况下最小的答案。
思路分析 ： 当 X 很大时，答案也会是一个很大的值，当 X 很小时，答案也会是一个很大的值，所以，只有当 X 适当时才会有最小的答案，那么这里显然就是三分，图像是一个凹函数，三分判断的条件就是一个求一个最大区间和就可以，因为区间绝对值和最大，它产生的情况有两种，一种区间的和是正值最大，一种区间的和是负值，让它最小，则绝对值会大。
代码示例 ：
#define ll long long
const int maxn = 2e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
double arr[maxn];
double pp[maxn];
int n;
double dp[maxn], dp2[maxn];
double ans = 1.0*inf;
double abc(double x){
return x < 0?-x:x;
}
double fun(double x){
for(int i = 1; i <= n; i++){
pp[i] = arr[i] - x;
}
memset(dp, 0, sizeof(dp));
memset(dp2, 0, sizeof(dp2));
double num = -1.0*inf;
for(int i = 1; i <= n; i++){
dp[i] = max(dp[i], dp[i-1]+pp[i]);
dp2[i] = min(dp2[i], dp2[i-1]+pp[i]);
double f = abc(dp[i]), f2 = abc(dp2[i]);
num = max(f, num);
num = max(f2, num);
}
return num;
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
cin >> n;
for(int i = 1; i <= n; i++){
scanf("%lf", &arr[i]);
}
double l = -20000, r = 20000;
for(int i = 1; i <= 100; i++){
double lm = l + (r-l)/3;
double rm = r - (r-l)/3;
double lf = fun(lm);
double rf = fun(rm);
if (lf > rf) {
l = lm;
ans = min(ans, rf);
}
else {
r = rm;
ans = min(ans, lf);
}
}
printf("%.15lf\n", ans);
return 0;
}
