NOIP提高组【JZOJ4809】挖金矿

Description

Data Constraint

Solution

这道题我们用二分答案就好。每次二分一个平均值，从1到n扫一下，选取一个总和比平均值个数大最多的数，若没有，则选择一个总和比平均值 个数相差最小的数即可。时间复杂度O(N*Hlog109)。

代码

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>#define ll long longusing namespace std;const int maxn=100005;ll a[maxn],n,m,i,t,j,k,num,p;double mid,l,r,b[maxn],c[maxn],sum,mx,x;bool bz;int main(){//  freopen("data.in","r",stdin);    scanf("%d%d",&n,&m);    for (i=1;i<=n;i++){        k=t+2;        for (j=1;j<=m;j++)            scanf("%lld",&a[++t]);        c[k-1]=b[k-1]=a[k-1];        for (j=k;j<=t;j++)            c[j]=b[j]=b[j-1]+a[j];        for (j=k-1;j<=t;j++)            c[j]/=j-k+2;    }    l=0;r=10000000000;    while (r-l>0.000001){        mid=(l+r+0.000001)/2;sum=0;num=0;        for (i=1;i<=n;i++){            mx=1000000000*n*m;            for (j=i*m;j>=(i-1)*m+1;j--)                if ((mid-c[j])*(j-(i-1)*m)<mx) mx=(mid-c[j])*(j-(i-1)*m),p=j;            sum+=b[p],num+=p-(i-1)*m;        }        x=sum/(num*1.0);        if (x>=mid) l=mid;        else r=mid-0.000001;    }    printf("%.4lf\n",l);}