Sabotage 二分

题目链接：http://www.usaco.org/index.php?page=viewproblem2&cpid=419

题意：有1,000,000个整数，要求从第2个到n-1个之间，删去一段连续的数，使得剩余数的平均数最小。（ 1<=所有数<= 1e6）.

题解：二分法，二分一个平均密度p，使原序列均减去p，求得2~n-1的最大连续和，可得到剩余数的和的正负值。若为正，则p小于mid；否则，p大于mid。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;const int N = 1e5+10;
const double eps = 1e-4;
double a[N],b[N],high,low,sum;
int n;double get(){double ans = 0,sum=0;double fumax = -1e8;int fl=1;for(int i=2;i<=n-1;i++){if(b[i]>0) fl=0;fumax = max(fumax,b[i]);}if(fl) return fumax;for(int i=2;i<=n-1;i++){if(sum+b[i]>0) sum += b[i];else sum = 0;ans = max(ans,sum);}return ans;
}double solve(){while(high-low>eps){double mid = (low+high)/2;for(int i=1;i<=n;i++)b[i] = a[i]-mid;double maxd = get();double tmp = sum-n*mid-maxd;if(tmp < 0) high = mid;else low = mid;}return low;
}int main(){
//    freopen("10.in","r",stdin);cin >> n;sum = 0;high = 1;low = 1e8;for(int i=1;i<=n;i++){scanf("%lf",&a[i]);sum += a[i];high = max(a[i],high);low = min(a[i],low);}double ans = solve();printf("%.3lf\n",ans);return 0;
}

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