这场可以说是相当水了。。然而一下涨了辣么多很高兴:)
这题题意看了很久。。楞是看不懂(怀疑自己和外国人的思维有些脱节?)。。然后还是别人告诉的题意。。
这题的话咋看无从下手,其实我们的目标就是要尽可能去掉多的数。一个数只能被整段区间去掉,而区间长度只能c的整数倍,因为再加上一两个元素没有作用,然后再想一下可以知道,如果2*c长度的区间分成2个长度为c的区间其实更合适,因为如果2个区间和并,次小值绝对不可能变大,相反可能会变小。。所以区间长度只能是c,或者干脆不给分区间。。。
这样一来,就可以先预处理区间最小数,很容易想到用单调队列,这里用了优先队列其实没有必要。。。
处理完后,就直接来一次dp就可以了
设扫到i位去掉数之和为d[i],区间右段为i时去掉的数为c[i]
d[i]=max(d[i-1],d[i-k]+c[i])
优先队列
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ > < ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-12
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 200005
#define nm 105
#define pi 3.1415926535897931
using namespace std;
const ll inf=1000000000;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
ll a[NM],c[NM],d[NM],ans,s;
ll n,k;
struct tmp{
int num;
bool operator<(const tmp&o)const{return a[num]>a[o.num];}
}b[NM];
priority_queue<tmp>q;
int main(){
n=read();k=read();
inc(i,1,n)b[i].num=i;
inc(i,1,n)a[i]=read();
inc(i,1,n)s+=a[i];
inc(i,1,k-1)q.push(b[i]);
inc(i,k,n){
q.push(b[i]);
tmp t=q.top();
while(t.num<=i-k)q.pop(),t=q.top();
c[i]=a[t.num];
}
inc(i,k,n)d[i]=max(d[i-1],d[i-k]+c[i]);
inc(i,1,n)ans=max(ans,d[i]);
printf("%I64d\n",s-ans);
return 0;
}
单调队列
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ > < ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-12
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 200005
#define nm 105
#define pi 3.1415926535897931
using namespace std;
const ll inf=1000000000;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
ll a[NM],c[NM],d[NM],ans,s;
int n,k,q[NM],qh,qt;
int main(){
n=read();k=read();k=min(k,n+1);
inc(i,1,n)a[i]=read();
inc(i,1,n)s+=a[i];
q[qh=qt=1]=1;
inc(i,2,k-1){while(a[i]<=a[q[qt]])qt--;q[++qt]=i;}
inc(i,k,n){
while(qh<=qt&&q[qh]<=i-k)qh++;
while(qh<=qt&&a[i]<=a[q[qt]])qt--;
q[++qt]=i;
c[i]=a[q[qh]];
}
inc(i,k,n)d[i]=max(d[i-1],d[i-k]+c[i]);
inc(i,1,n)ans=max(ans,d[i]);
printf("%I64d\n",s-ans);
return 0;
}
E. Cashback
time limit per test
memory limit per test
input
output
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the
smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Copy
3 51 2 3
Output
6
Copy
12 101 1 10 10 10 10 10 10 9 10 10 10
Output
92
Copy
7 22 3 6 4 5 7 1
Output
17
Copy
8 41 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10]
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1]
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1]