Problem - 1796D - Codeforces
思路：想了个假dp做法推了半天，果然是dp。考虑用dp[i][j]表示以i结尾的，并且选择j个＋x的最长连续子序列，那么如果我不选择第i位，那么会有f[i][j]=max(w[i]-x,f[i-1][j]+w[i]-x)，同时i>j也要满足，因为如果i<=j，那么所有的我必须都要选择，如果第i位我选择，那么会有
f[i][j]=max(w[i]+x,f[i-1][j-1]+w[i]+x)，同时要满足j-1>=0另外我们还要保证我们总共要选择k个，在前i个中我们选择了j个，那么要满足在n-i中至少选择k-j个，即要满足n-i>=k-j，所以j>=k-n+i，所以我们只要算所有的f[i][j]同时取一个max即可，因为我们保证了我们枚举到的i,j都满足总共要选择k个的条件
// Problem: D. Maximum Subarray
// Contest: Codeforces - Educational Codeforces Round 144 (Rated for Div. 2)
// URL: https://codeforces.com/problemset/problem/1796/D
// Memory Limit: 512 MB
// Time Limit: 2000 ms#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<bitset>
#include<deque>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#include<set>
#include<cstdlib>
#define fi first
#define se second
#define i128 __int128
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> PII;
typedef pair<int,pair<int,int> > PIII;
const double eps=1e-7;
const int N=5e5+7 ,M=5e5+7, INF=0x3f3f3f3f,mod=1e9+7,mod1=998244353;
const long long int llINF=0x3f3f3f3f3f3f3f3f;
inline ll read() {ll x=0,f=1;char c=getchar();while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
while(c>='0'&&c<='9') {x=(ll)x*10+c-'0';c=getchar();} return x*f;}
inline void write(ll x) {if(x < 0) {putchar('-'); x = -x;}if(x >= 10) write(x / 10);putchar(x % 10 + '0');}
inline void write(ll x,char ch) {write(x);putchar(ch);}
void stin() {freopen("in_put.txt","r",stdin);freopen("my_out_put.txt","w",stdout);}
bool cmp0(int a,int b) {return a>b;}
template<typename T> T gcd(T a,T b) {return b==0?a:gcd(b,a%b);}
template<typename T> T lcm(T a,T b) {return a*b/gcd(a,b);}
void hack() {printf("\n----------------------------------\n");}int T,hackT;
int n,m,k;
int w[N];
ll f[N][21];void solve() {n=read(),k=read();int x=read();for(int i=1;i<=n;i++) w[i]=read();for(int i=1;i<=n;i++) {for(int j=0;j<=k;j++) {f[i][j]=-llINF;}}ll res=0;for(int i=1;i<=n;i++) {for(int j=max(0,k+i-n);j<=min(i,k);j++) {if(j<i) f[i][j]=max((ll)w[i]-x,f[i-1][j]+w[i]-x);if(j>0) f[i][j]=max(f[i][j],max((ll)w[i]+x,f[i-1][j-1]+w[i]+x));f[i][j]=max(f[i][j],0ll);res=max(res,f[i][j]);}}printf("%lld\n",res);
} int main() {// init();// stin();scanf("%d",&T);// T=1; while(T--) hackT++,solve();return 0;
}
参考博客：https://www.cnblogs.com/onlyblues/p/17177714.html
这个博客讲的挺详细的
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