USACO14MAR Sabotage

题目地址

题解

二分答案，设为$ans$，那么判定条件

$$C(ans):\exists [i,j],\frac {sum[n]-sum[i...j]}{n-(j-i+1)}\le ans$$

移项，等价于

$$sum[n]-ans\times n\le sum[j]-ans\times j-(sum[i-1]-ans\times (i-1))$$

求右边的最大值即可。

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cctype>
#define INF 1000000000
using namespace std;
typedef long long ll;
int read(){
    int f=1,x=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}
    while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
    return f*x; 
}
int n,a[100005];
double eps=1e-4,sum[100005]={0}; 
bool C(double ans){
    double mini=sum[1]-ans,maxi=sum[2]-2*ans-(sum[1]-ans);
    for(int i=2;i<n;i++){
        maxi=max(maxi,sum[i]-(double)i*ans-mini);
        mini=min(mini,sum[i]-(double)i*ans);
    }
    return (sum[n]-(double)n*ans<=maxi);
}
void init(){
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read(),sum[i]=sum[i-1]+a[i];
}
void solve(){
    double L=0,R=(a[1]+a[n])*0.5,M;
    while(R-L>eps){
        M=(R+L)*0.5;
        if(C(M))R=M;else L=M;
    }
    printf("%.3lf\n",L);
}
int main(){
    init();
    solve();
    return 0;
}