NOIP 模拟 $28; m 客星璀璨之夜$

题解 (by;zjvarphi)

概率与期望，考虑 ( m dp)

设 (dp_{i,j}) 为消除 (i～j) 这一段行星的期望，转移：

枚举 (k) 为当前状态下第一个撞击的行星，分向左，向右。

[ m dp_{i,j}=sum_{k=i}^jdp_{i,k-1}+dp_{k+1,i}+E_{k+1,j}-pos_k ]

[ m dp_{i,j}=sum_{k=i}^jdp_{i,k-1}+dp_{k+1,i}-E_{i,k-1}+pos_k ]

( m E_{i,j}) 表示 ( m i->j) 的期望位置，转移与 ( m dp) 类似。

这样就可以 (mathcal O m (n^3))，加个前缀和即可优化成 (mathcal O m(n^2))

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf,OPUT[100];
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++;
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
    template<typename T>inline void print(T x,char t) {
        if (x<0) putchar('-'),x=-x;
        if (!x) return putchar('0'),(void)putchar(t);
        ri cnt(0);
        while(x) OPUT[p(cnt)]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
        return (void)putchar(t);
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #define int long long
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=3e3+7,MOD=998244353; 
    int dp[N][N],suf[N][N],pre[N][N],ex[N][N],a1[N][N],a2[N][N],inv[N],pa[N],n;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        read(n);
        inv[1]=1;
        ri al=(n<<1)+1;
        for (ri i(2);i<=al;p(i)) inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;
        for (ri i(1),pla;i<=n;p(i)) {
            read(ex[i][i-1]),read(pla);
            pa[i]=(pa[i-1]+pla)%MOD;
        }
        read(ex[n+1][n]);
        for (ri i(n);i;--i)
            for (ri j(i);j<=n;p(j)) {
                pre[i][j]=(pre[i][j-1]+ex[i][j-1])%MOD;
                suf[j][i]=(suf[j][i+1]+ex[i+1][j])%MOD;
                ex[i][j]=(pre[i][j]+suf[j][i])*inv[j-i+1]%MOD*inv[2]%MOD;
                dp[i][j]=(dp[i][j]+2*(a1[i][j-1]+a2[j][i+1])%MOD)%MOD;
                dp[i][j]=(dp[i][j]+pa[j]-pa[i-1]-pre[i][j])%MOD;
                dp[i][j]=(dp[i][j]-pa[j]+pa[i-1]+suf[j][i])%MOD;
                dp[i][j]=dp[i][j]*inv[j-i+1]%MOD*inv[2]%MOD;
                a1[i][j]=(a1[i][j-1]+dp[i][j])%MOD;
                a2[j][i]=(a2[j][i+1]+dp[i][j])%MOD;    
            }
        print((dp[1][n]+MOD)%MOD,'
');
        return 0;
    }
    #undef int
}
int main() {return nanfeng::main();}