4 summer exams: felling cut (block divisible)

 

topic:

analysis:

A look at this question, eh not to go half a d O (n) to check can not it do?

But in fact it is not dichotomous! ! Because not satisfied monotonic. For example, this set of data: a [i] = 4 k = 1 if d = 3, the will cut on day 6, cost = 2, d = 4, the will cut on day 4, cost = 0, apparent d more big, but the better! ! !

So half before it must consider whether monotone , since that when met, should also be a little counter-examples (This question is inside a small d is not met, but the big meet).

So not satisfied with how to do it -> descending enumeration, enumeration to meet on the first break away. D is the upper bound min {a [i]} + k (since k> = d-minn)

Of course, it will be T out.

So hand push equation:

And ai / d (the rounding) will find a duplicate, then consider divisible block (generally the block is divisible by the proof ): How divisible block

This question is divisible by the block to be proved. . .

#include<bits/stdc++.h>
using namespace std;
#define inf 2100000000
#define N 105
#define ll long long
ll k,minn=inf,ans,a[N],sum;
int n;
bool fl[N];
int main()
{
    freopen("cut.in","r",stdin);
    freopen("cut.out","w",stdout);
    scanf("%d%lld",&n,&k);
    sum=k;
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]),sum+=a[i],minn=min(minn,a[i]);
    minn+=k;
    ll d=1;
    while(1){
        ll tot=0;
        for(int i=1;i<=n;i++){
            if(a[i]%d) tot+=(a[i]/d +1)*d;//讨论能不能整除 从而向上取整 
            else tot+=a[i]; 
        }
        if(tot<=sum) ans=d;
        if(d>=minn) break;
        d=sum/(sum/(d+1));
    }
    printf("%lld\n",ans);
    return 0;
}