題意:
每個數都減x,找到合适的x,使得最大的abs(連續子序列的和)->最小
解法:
三分,類開口朝上的二次函數
執行個體代碼:
#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <vector>
#define rep(i,n) for(i=0;i<n;i++)
#define cle(x) memset(x,0,sizeof(x))
#define ll long long
const int maxn=+;
using namespace std;
ll a[maxn];
long double b[maxn];
long double ans;
int n;
long double doit (long double x){
for(int i=;i<=n;i++)b[i]=a[i];
for(int i=;i<=n;i++)b[i]-=(x*i);
long double mi=;
long double ma=;
long double re=;
for(int i=;i<=n;i++){
re=max(re,max(abs(b[i]-mi),abs(b[i]-ma)));
mi=min(mi,b[i]);
ma=max(ma,b[i]);
}
ans=min(ans,re);
return re;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
while(cin>>n){
int i;
for(i=;i<=n;i++)scanf("%I64d",&a[i]);
for(i=;i<=n;i++)a[i]=a[i-]+a[i];
long double l=-,r=;
int T=;
ans=;
while(T--){
long double ml=l+(r-l)/;
long double mr=r-(r-l)/;
if(doit(ml)>doit(mr))l=ml;
else r=mr;
}
printf("%.12lf\n",(double)ans);
}
return ;
}