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 long
using 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);
}