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  • Solution -「BalticOI 2004」Sequence

    Description

    Link.

    Given is a sequencen (A) of (n) intergers.

    Construct a stricly increasing sequence (B) of (n) intergers that makes the sum of (|B_{i}-A_{i}|) the smallest.

    Solution

    First, we make (a_{i}:=a_{i}-i). Then we just make "strictly increasing" become "unstrictly increasing".

    1. for (a_{1}le a_{2}lecdotsle a_{n}):

      When (B) is the same as (A), it gets the minimum answer.

    2. for (a_{1}ge a_{2}gecdotsge a_{n}):

      When for each (i), (B_{i}=A_{lfloorfrac{n}{2} floor}), it gets the minimum answer.

    Maybe we can divide (A) into m parts.

    Such like: ([l_{1},r_{1}],cdots,[l_{m},r_{m}])

    that satisfies: for each (i), sequence (A[l_{i},r_{i}]) is unstrictly increasing/decreasing.

    So we can get the answer to each interval.

    Let's consider how to merge the answers.

    When we're merging two intervals, we can just get the new median of the two intervals.


    So things above are just bullshit.

    Parallel Searching!

    FUCK YOU.

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    const LL INF=1e18;
    int n;
    LL a[1000010],b[1000010],ans;
    void solve(LL l,LL r,int fr,int ba)
    {
    	if(l>=r||fr>ba)	return;
    	LL mid=(l+r)>>1,tmp=0,mn=INF,pos=0;
    	for(int i=fr;i<=ba;++i)	tmp+=abs(a[i]-mid-1);
    	mn=tmp,pos=fr-1;
    	for(int i=fr;i<=ba;++i)
    	{
    		tmp-=abs(a[i]-mid-1);
    		tmp+=abs(a[i]-mid);
    		if(tmp<mn)	mn=tmp,pos=i;
    	}
    	for(int i=fr;i<=pos;++i)	b[i]=mid;
    	for(int i=pos+1;i<=ba;++i)	b[i]=mid+1;
    	solve(l,mid,fr,pos),solve(mid+1,r,pos+1,ba);
    }
    int main()
    {
    	scanf("%d",&n);
    	for(int i=1;i<=n;++i)	scanf("%lld",&a[i]),a[i]-=i;
    	solve(-INF,INF,1,n);
    	for(int i=1;i<=n;++i)	ans+=abs(a[i]-b[i]);
    	printf("%lld
    ",ans);
    	for(int i=1;i<=n;++i)	printf("%lld ",b[i]+i);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/orchid-any/p/14490695.html
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