这是我赛后抄的写的代码。

#include <bits/stdc++.h>
#define lb(x) (x&-x)
#define L(i,j,k) for(int i=(j);i<=(k);++i)
#define R(i,j,k) for(int i=(j);i>=(k);--i)
#define swap(a,b) (a^=b^=a^=b)
#define int long long

using namespace std;
using i64 = long long;

typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
void chmin(int &x, int c) {
    x = min(x, c);
}
void chmax(int &x, int c) {
    x = max(x, c);
}

namespace fast_IO{
#define IOSIZE 100000
int precision=3,POW[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};char ibuf[IOSIZE],obuf[IOSIZE],*p1=ibuf,*p2=ibuf,*p3=obuf;
#ifdef ONLINE_JUDGE
#define getchar()((p1==p2)and(p2=(p1=ibuf)+fread(ibuf,1,IOSIZE,stdin),p1==p2)?(EOF):(*p1++))
#define putchar(x)((p3==obuf+IOSIZE)&&(fwrite(obuf,p3-obuf,1,stdout),p3=obuf),*p3++=x)
#define isdigit(ch)(ch>47&&ch<58)
#define isspace(ch)(ch<33)
#endif
template<typename T>inline T read(){T s=0;int w=1;char ch;while(ch=getchar(),!isdigit(ch)&&(ch!=EOF))if(ch==45)w=-1;if(ch==EOF)return 0;while(isdigit(ch))s=s*10+ch-48,ch=getchar();return s*w;}template<typename T>inline bool read(T&s){s=0;int w=1;char ch;while(ch=getchar(),!isdigit(ch)&&(ch!=EOF))if(ch==45)w=-1;if(ch==EOF)return 0;while(isdigit(ch))s=s*10+ch-48,ch=getchar();return s*=w,1;}inline bool read(char&s){while(s=getchar(),isspace(s));return 1;}inline bool read(char*s){char ch;while(ch=getchar(),isspace(ch));if(ch==EOF)return 0;while(!isspace(ch))*s++=ch,ch=getchar();*s='\000';return 1;}template<typename T>inline void print(T x){if(x<0)putchar(45),x=-x;if(x>9)print(x/10);putchar(x%10+48);}inline void print(char x){putchar(x);}inline void print(char*x){while(*x)putchar(*x++);}inline void print(const char*x){for(int i=0;x[i];i++)putchar(x[i]);}inline bool read(std::string&s){s="";char ch;while(ch=getchar(),isspace(ch));if(ch==EOF)return 0;while(!isspace(ch))s+=ch,ch=getchar();return 1;}inline void print(std::string x){for(int i=0,n=x.size();i<n;i++)putchar(x[i]);}inline bool read(bool&b){char ch;while(ch=getchar(),isspace(ch));b=ch^48;return 1;}inline void print(bool b){putchar(b+48);}inline bool read(double&x){int a=0,b=0;char ch=getchar();bool f=0;while(ch<48||ch>57){if(ch==45)f=1;ch=getchar();}while(47<ch&&ch<58){a=(a<<1)+(a<<3)+(ch^48);ch=getchar();}if(ch!=46){x=f?-a:a;return 1;}ch=getchar();while(47<ch&&ch<58){b=(b<<1)+(b<<3)+(ch^48),ch=getchar();}x=b;while(x>1)x/=10;x=f?-a-x:a+x;return 1;}inline void print(double x){if(x<0){putchar(45),x=-x;}x=round((long double)x*POW[precision])/POW[precision],print((long long)x),putchar(46),x-=(long long)(x);for(int i=1;i<=precision;i++)x*=10,putchar(x+48),x-=(int)x;}template<typename T,typename...T1>inline int read(T&a,T1&...other){return read(a)+read(other...);}template<typename T,typename...T1>inline void print(T a,T1...other){print(a),print(other...);}struct Fast_IO{~Fast_IO(){fwrite(obuf,p3-obuf,1,stdout);}}io;template<typename T>Fast_IO&operator>>(Fast_IO&io,T&b){return read(b),io;}template<typename T>Fast_IO&operator<<(Fast_IO&io,T b){return print(b),io;}
#define cout io
#define cin io
#define endl '\n'
}using namespace fast_IO;
const int maxn = 3e5 + 10, mod = 998244353;
int dp[110][10010];
int A[maxn], N, K, X, sum;
vector<int> g;

void solve() {
    cin >> N >> K >> X;
    L (i, 1, N) cin >> A[i], sum += A[i];
    g.push_back(0);
    L (i, 1, N) L (j, 0, N - 1)
        g.push_back(A[i] + j * X);
    sort (g.begin(), g.end());
    g.erase(unique(g.begin(), g.end()), g.end());
    memset (dp, 0x3f, sizeof dp);
    int M = g.size() - 1;
    dp[0][0] = 0;
    for (int i = 1; i <= M; i ++) {
        dp[0][i] = 0;
    }
    for (int i = 1; i <= N; i ++ ) {
        int x = lower_bound(g.begin(), g.end(), A[i]) - g.begin();
        for (int j = x; j <= M; j ++ ) {
            dp[i][j] = dp[i][j - 1];
            for (int k = max (0ll, i - K); k < i; k ++) {
                chmin (dp[i][j], dp[k][upper_bound(g.begin(), g.end(), g[j] - X) - g.begin() - 1] + (i - k) * g[j]);
            }
        }
    }
    cout << dp[N][M] - sum << '\n';
}

signed main() {
    int T = 1;
    while (T--)solve();
    return 0;
}

这个过了

为什么把

L (i, 1, N) L (j, 0, N - 1)
        g.push_back(A[i] + j * X);

中的 N - 1 改为 N 就过不了了？？？我很不理解