PKUWC2018 Slay the Spire

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枚举抽出强化牌的张数x。

然后最优策略大概是先从大到小打出一些强化牌，再从大到小打出一些攻击牌。

显然只要有攻击牌，一定会攻击至少一次。

由于强化牌至少攻击翻倍，所以在前一条的基础上显然强化牌打得越多越好。

所以说随着x增加，强化牌和攻击牌打出的张数如下表所示（当然如果牌数不够那也没办法）：

然后直接计数。

/*
Author: CNYALI_LK
LANG: C++
PROG: 2538.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
    const int SIZE = (1 << 21) + 1;
    char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
    // getchar
    #define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
    // print the remaining part
    inline void flush () {
        fwrite (obuf, 1, oS - obuf, stdout);
        oS = obuf;
    }
    // putchar
    inline void putc (char x) {
        *oS ++ = x;
        if (oS == oT) flush ();
    }
    // input a signed integer
    inline void read (signed &x) {
        for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
    }

    inline void read (long long &x) {
        for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
    }
    inline void read (char &x) {
        x=gc();
    }
    inline void read(char *x){
        while((*x=gc())=='\n' || *x==' '||*x=='\r');
        while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
    }
    template<typename A,typename ...B>
    inline void read(A &x,B &...y){
        read(x);read(y...);
    }
    // print a signed integer
    inline void write (signed x) {
        if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
        while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
        while (qr) putc (qu[qr --]);
    }

    inline void write (long long x) {
        if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
        while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
        while (qr) putc (qu[qr --]);
    }
    inline void write (char x) {
        putc(x);
    }
    inline void write(const char *x){
        while(*x){putc(*x);++x;}
    }
    inline void write(char *x){
        while(*x){putc(*x);++x;}
    }
    template<typename A,typename ...B>
    inline void write(A x,B ...y){
        write(x);write(y...);
    }
    //no need to call flush at the end manually!
    struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int f[3005][3005],fcnt[3005],g[3005][3005],gcnt[3005],c[3005][3005],w1[3005],w2[3005];
const int p=998244353;
int sum(int a,int b){return (a+=b)>=p?a-p:a;}
int main(){
#ifdef cnyali_lk
    freopen("2538.in","r",stdin);
    freopen("2538.out","w",stdout);
#endif
    int n,m,k,t;
    read(t);
    for(int i=0;i<=3000;++i){
        c[i][0]=c[i][i]=1;
        for(int j=1;j<i;++j)c[i][j]=sum(c[i-1][j],c[i-1][j-1]);
    }
    for(;t;--t){
        read(n,m,k);
        for(int i=1;i<=n;++i)read(w1[i]);
        sort(w1+1,w1+n+1);
        for(int i=1;i<=n;++i)read(w2[i]);
        sort(w2+1,w2+n+1);
        for(int i=1;i<=n;++i)f[0][i]=0;
        f[0][n+1]=1;
        for(int i=1;i<=k-1;++i){
            f[i][n+1]=0;
            fcnt[i-1]=f[i-1][n+1];
            for(int j=n;j;--j){
                f[i][j]=(ll)fcnt[i-1]*w1[j]%p;    
                fcnt[i-1]=sum(fcnt[i-1],f[i-1][j]);
            }
        }
        fcnt[k-1]=f[k-1][n+1];
        for(int i=1;i<=n;++i)fcnt[k-1]=sum(fcnt[k-1],f[k-1][i]);
        for(int i=k;i<=m;++i){
            fcnt[i]=0;

            for(int j=1;j<=n+1;++j){
                f[i][j]=(ll)f[k-1][j]*c[j-1][i-k+1]%p;
                fcnt[i]=sum(fcnt[i],f[i][j]);
            }
        }
        for(int i=1;i<=n;++i)g[k-1][i]=w2[i];
        
        for(int i=k-2;~i;--i){
            gcnt[i+1]=0;
            for(int j=n;j;--j){
                g[i][j]=sum(gcnt[i+1],(ll)w2[j]*c[n-j][k-i-1]%p);
                gcnt[i+1]=sum(gcnt[i+1],g[i+1][j]);
            }
        }
        for(int i=1;i<=n;++i)gcnt[0]=sum(gcnt[0],f[1][i]);
        for(int i=k;i<=m;++i){
            gcnt[i]=0;
            for(int j=1;j<=n;++j){
                g[i][j]=(ll)g[k-1][j]*c[j-1][m-i-1]%p;    
                gcnt[i]=sum(gcnt[i],g[i][j]);
            }
        }
        for(int i=0;i<k;++i){
            gcnt[i]=0;
            for(int j=1;j<=n;++j){
                g[i][j]=(ll)g[i][j]*c[j-1][m-k]%p;
                gcnt[i]=sum(gcnt[i],g[i][j]);
            }
        }    
        int ans=0;
        for(int i=0;i<=m;++i){
//            printf("%d,%d\n",fcnt[i],gcnt[i]);
            ans=sum(ans,(ll)fcnt[i]*gcnt[i]%p);
        }
        printf("%d\n",ans);
    }
    return 0;
}