NOIP 模拟 $25\; \rm random$

题解 \(by\;zj\varphi\)

期望好题。

通过推规律可以发现每个逆序对的贡献都是 \(1\)，那么在所有排列中有多少逆序对，贡献就是多少。

\[\rm num_i=(i-1)!\sum_{j=1}^{i-1}j+i*num_{i-1}\\
\]

最后化减完可以得到

\[\rm ans=\frac{\sum_{i=1}^n\frac{i*(i-1)}{3}}{n}\\
ans=\frac{n^2-1}{9}
\]

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;
    #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;
    }
    char OPUT[100];
    template<typename T>inline void print(T x){
        if(x<0) putchar('-'),x=-x;
        if(!x) return putchar('0'),(void)putchar('\n');
        ri cnt=0;
        while(x) OPUT[++cnt]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]+'0');
        return (void)putchar('\n');
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #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;}
    typedef long long ll;
    static const int MOD=998244353;
    int T,inv;
    ll n;
    inline int fpow(int x,int y) {
        int res=1;
        while(y) {
            if (y&1) res=1ll*res*x%MOD;
            x=1ll*x*x%MOD;
            y>>=1;
        }
        return res;
    }
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        read(T);
        inv=fpow(9,MOD-2);
        for (ri z(1);z<=T;p(z)) {
            read(n);
            n%=MOD;
            print((n*n%MOD-1ll)*inv%MOD);
        }
        return 0;
    }
}
int main() {return nanfeng::main();}