「CF1197D」Yet Another Subarray Problem【分治或dp】

D. Yet Another Subarray Problem

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

You are given an array $a_1,a_2,\dots ,a_n$ and two integers ? and ?.

You can choose some subarray $a_l,a_{l+1},\dots ,a_{r-1},a_r$.

The cost of subarray $a_l,a_{l+1},\dots ,a_{r-1},a_r$ is equal to ${\sum }_{i=l}^r{a}_i-k\times \lceil \frac{r-l+1}{m}\rceil$,where $\lceil x\rceil$ is the least integer greater than or equal to $x$.

The cost of empty subarray is equal to zero.

For example, if $m=3,k=10$ and $a=[2,-4,15,-3,4,8,3]$, then the cost of some subarrays are:

$a_3\dots a_3:15-k\lceil \frac{1}{3}\rceil =15-10=5;$
$a_3\dots a_4:(15-3)-\lceil \frac{1}{3}\rceil =12-10=2;$
$a_3\dots a_5:(15-3+4)-\lceil \frac{1}{3}\rceil =16-10=6;$
$a_3\dots a_6:(15-3+4+8)-\lceil \frac{1}{3}\rceil =24-20=4;$
$a_3\dots a_7:(15-3+4+8+3)-\lceil \frac{1}{3}\rceil =27-20=7.$
Your task is to find the maximum cost of some subarray (possibly empty) of array $a$.

Input

The first line contains three integers $n,m,andk(1\le n\le 3\times 10^5,1\le m\le 10,1\le k\le 10^9)$.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n(-10^9\le a_i\le 10^9)$.

Output

Print the maximum cost of some subarray of array $a$.

Examples

input

7 3 10
2 -4 15 -3 4 8 3

output

7

input

5 2 1000
-13 -4 -9 -20 -11

output

0

题意

• 给你一个数组，以及给你$m,k$,求下面这个式子的最大值
  $$\sum _{i=l}^r{a}_i-k\times \lceil \frac{r-l+1}{m}\rceil$$

题解

• 由于$1\le m\le 10$，可以考虑分治解决这个问题，分治合并的时候，先处理左区间每一个点的后缀和，然后计算后缀对应区间的长度$\%m$意义下的上式最大值，然后扫右边的时候，枚举左边（也就是区间$[l,mid]$）区间的长度$\%m$意义下的长度，取个$max$就行了，时间复杂度$O(nm\log n)$
• 还可以考虑$dp$解决这个问题，就是定义$dp[i][j]$表示以i为右端点，左端点$l$满足$(r-l+1)\%m=j$的上述函数最大值，转移的时候考虑从右侧第一个$m$分段内转移还是$i-m$前转移取最大就行了

分治解法代码

#include
 
using namespace std;
const int maxn=3e5+10;
 
int n,m;
long long a[maxn],k;
long long cur[100],sum[maxn];
 
int solve(int b)
{
    return b%m==0?b/m:b/m+1;
}
 
long long dfs(int l,int r)
{
    if(l==r) return a[l]-k;
    int mid=(l+r)>>1;long long res=-1e18;
    for(int i=0;i<m;i++) cur[i]=-1e18;sum[mid+1]=0;
    for(int i=mid;i>=l;i--) {
        sum[i]=a[i]+sum[i+1];
        cur[(mid-i+1)%m]=max(cur[(mid-i+1)%m],sum[i]-k*solve(mid-i+1));
    }
    sum[mid]=0;
    for(int i=mid+1;i<=r;i++) {
        int now=(i-mid)%m;sum[i]=sum[i-1]+a[i];
        for(int j=0;j<m;j++){
            if(j==0) res=max(res,cur[j]+sum[i]-k*solve(i-mid));
            else {
                if(j+now>m) res=max(res,cur[j]+sum[i]-k*solve(i-mid));
                else res=max(res,cur[j]+sum[i]-k*((i-mid)/m));
            } 
        }
    }
 
    return max(res,max(dfs(l,mid),dfs(mid+1,r)));
}
 
int main()
{
    scanf("%d %d %lld",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    printf("%lld\n",max(0LL,dfs(1,n)));
}

dp代码

#include

using namespace std;
const int maxn=3e5+10;

int n,m,k,a[maxn];
long long dp[maxn][12],sum[maxn];

int main()
{
    scanf("%d %d %d",&n,&m,&k);long long ans=0;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),sum[i]=sum[i-1]+a[i];
    for(int i=1;i<=n;i++) {
        if(i-m>=0) dp[i][0]=max(dp[i][0],dp[i-m][0]+sum[i]-sum[i-m]-k);
        for(int j=1;j<m;j++) {
            if(i-j>=0) dp[i][j]=max(dp[i][j],sum[i]-sum[i-j]-k);
            if(i>=m) dp[i][j]=max(dp[i][j],sum[i]-sum[i-m]-k+dp[i-m][j]);
        }
        for(int j=0;j<m;j++) ans=max(ans,dp[i][j]);
    }
    printf("%lld\n",ans);
}