有程序员能评价一下竞赛人员的代码吗

#include <bits/stdc++.h>//一道线段树题目，还用到了离散化
#define int long long
using namespace std;
const int M = 998244353, N = 2e5 + 10;
struct segmenttree
{
    int l, r, x, lazy_jia, lazy_cheng, sum;
} a[N * 4];
struct node
{
    int l, r, size;
} ss[N];
int m, n, b[N], ls[N];
map<int, int> p;
void write(int x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x >= 10)
    {
        write(x / 10);
    }
    putchar(x % 10 + '0');
}
inline int read()
{
    char ch = getchar();
    int ret = 0, f = 1;
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
        {
            f = -1;
        }
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = (ret << 1) + (ret << 3) + ch - '0';
        ch = getchar();
    }
    return ret * f;
}
void js(int l, int r, int x)
{
    a[x].l = l, a[x].r = r, a[x].lazy_cheng = 1;
    if (l == r)
    {
        return;
    }
    int mid = (l + r) / 2;
    js(l, mid, x * 2);
    js(mid + 1, r, x * 2 + 1);
    a[x].sum = (a[x * 2].sum + a[x * 2 + 1].sum) % M;
}
inline void downop(int x)
{
    a[x * 2].sum *= a[x].lazy_cheng;
    a[x * 2].sum %= M;
    a[x * 2 + 1].sum *= a[x].lazy_cheng;
    a[x * 2 + 1].sum %= M;
    a[x * 2].sum += a[x].lazy_jia % M * (ss[a[x * 2].r].r - ss[a[x * 2].l].l + 1) % M;
    a[x * 2].sum %= M;
    a[x * 2 + 1].sum += a[x].lazy_jia % M * (ss[a[x * 2 + 1].r].r - ss[a[x * 2 + 1].l].l + 1) % M;
    a[x * 2 + 1].sum %= M;
    a[x * 2].lazy_cheng *= a[x].lazy_cheng;
    a[x * 2].lazy_cheng %= M;
    a[x * 2 + 1].lazy_cheng *= a[x].lazy_cheng;
    a[x * 2 + 1].lazy_cheng %= M;
    a[x * 2].lazy_jia *= a[x].lazy_cheng;
    a[x * 2].lazy_jia %= M;
    a[x * 2 + 1].lazy_jia *= a[x].lazy_cheng;
    a[x * 2 + 1].lazy_jia %= M;
    a[x * 2].lazy_jia += a[x].lazy_jia;
    a[x * 2].lazy_jia %= M;
    a[x * 2 + 1].lazy_jia += a[x].lazy_jia;
    a[x * 2 + 1].lazy_jia %= M;
    a[x].lazy_cheng = 1;
    a[x].lazy_jia = 0;
}
void jia(int l, int r, int x, int s)
{
    if (l > r)
        return;
    if (a[x].l >= l && a[x].r <= r)
    {
        a[x].lazy_jia += s;
        a[x].lazy_jia %= M;
        a[x].sum += s % M * (ss[a[x].r].r - ss[a[x].l].l + 1) % M;
        a[x].sum %= M;
        return;
    }
    downop(x);
    int mid = (a[x].l + a[x].r) / 2;
    if (mid >= l)
    {
        jia(l, r, x * 2, s);
    }
    if (mid < r)
    {
        jia(l, r, x * 2 + 1, s);
    }
    a[x].sum = (a[x * 2].sum + a[x * 2 + 1].sum) % M;
}
int he(int l, int r, int x)
{
    if (l > r)
        return 0;
    if (a[x].l >= l && a[x].r <= r)
    {
        return a[x].sum %= M;
    }
    downop(x);
    int mid = (a[x].l + a[x].r) / 2, ret = 0;
    if (mid >= l)
    {
        ret += he(l, r, x * 2) % M;
    }
    if (mid < r)
    {
        ret += he(l, r, x * 2 + 1) % M;
        ret %= M;
    }
    return ret % M;
}
void cheng(int l, int r, int x, int s)
{
    if (l > r)
        return;
    if (a[x].l >= l && a[x].r <= r)
    {
        a[x].sum *= s;
        a[x].sum %= M;
        a[x].lazy_cheng *= s;
        a[x].lazy_cheng %= M;
        a[x].lazy_jia *= s;
        a[x].lazy_jia %= M;
        return;
    }
    downop(x);
    int mid = (a[x].l + a[x].r) / 2;
    if (mid >= l)
        cheng(l, r, x * 2, s);
    if (mid < r)
        cheng(l, r, x * 2 + 1, s);
    a[x].sum = (a[x * 2].sum + a[x * 2 + 1].sum) % M;
}
signed main()
{
    ios::sync_with_stdio(0);
    n = read();
    for (int i = 1; i <= n; i++)
    {
        b[i] = read();
        ls[i] = b[i];
    }
    stable_sort(ls + 1, ls + n + 1);
    m = unique(ls + 1, ls + n + 1) - ls - 1;
    ss[1].l = 1, ss[1].r = ls[1], ss[1].size = ls[1];
    for (int i = 2; i <= m; i++)
    {
        ss[i].l = ls[i - 1] + 1;
        ss[i].r = ls[i];
        ss[i].size = ls[i] - ls[i - 1];
    }
    for (int i = 1; i <= m; i++)
    {
        p[ls[i]] = i;
    }
    js(1, m, 1);
    jia(1, p[b[1]], 1, 1);
    for (int i = 2; i <= n; i++)
    {
        int s = he(1, m, 1);
        cheng(p[b[i]] + 1, m, 1, 0);
        cheng(1, p[b[i]], 1, -1);
        jia(1, p[b[i]], 1, s);
    }
    write((he(1, m, 1) + M) % M);
    return 0;
}