take 牛客网多校

https://www.nowcoder.com/acm/contest/143/F

第i个钻石被拿<=>[1,i-1]中所有比第i个钻石大的钻石都没被拿 算出这个概率 乘上次数1 就是期望 累加即可

菜的一批

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define M 998244353
 
struct node
{
    int l;
    int r;
    ll val;
};
 
node tree[400010];
ll p[100010];
int d[100010],tmp[100010];
int n,len;
 
ll quickpow(ll a,ll b)
{
    ll res;
    res=1;
    while(b>0)
    {
        if(b%2) res=(res*a)%M;
        a=(a*a)%M,b/=2;
    }
    return res;
}
 
void pushup(int cur)
{
    tree[cur].val=(tree[2*cur].val*tree[2*cur+1].val)%M;
}
 
void build(int l,int r,int cur)
{
    int m;
    tree[cur].l=l;
    tree[cur].r=r;
    tree[cur].val=1;
    if(l==r) return;
    m=(l+r)/2;
    build(l,m,2*cur);
    build(m+1,r,2*cur+1);
}
 
void update(int tar,ll val,int cur)
{
    if(tree[cur].l==tree[cur].r)
    {
        tree[cur].val=(tree[cur].val*val)%M;
        return;
    }
    if(tar<=tree[2*cur].r) update(tar,val,2*cur);
    else update(tar,val,2*cur+1);
    pushup(cur);
}
 
ll query(int pl,int pr,int cur)
{
    ll res;
    if(pl<=tree[cur].l&&tree[cur].r<=pr) return tree[cur].val;
    res=1;
    if(pl<=tree[2*cur].r) res=(res*query(pl,pr,2*cur))%M;
    if(pr>=tree[2*cur+1].l) res=(res*query(pl,pr,2*cur+1))%M;
    return res;
}
 
int main()
{
    ll ny,ans,res;
    int i,pos;
    scanf("%d",&n);
    ny=quickpow(100,M-2);
    for(i=1;i<=n;i++)
    {
        scanf("%lld%d",&p[i],&d[i]);
        p[i]=(p[i]*ny)%M;
        tmp[i]=d[i];
    }
    sort(tmp+1,tmp+n+1);
    len=unique(tmp+1,tmp+n+1)-tmp-1;
    build(1,len,1);
    ans=0;
    for(i=1;i<=n;i++)
    {
        pos=lower_bound(tmp+1,tmp+len+1,d[i])-tmp;
        if(pos+1<=len) ans=(ans+(p[i]*query(pos+1,len,1))%M)%M;
        else ans=(ans+p[i])%M;
        update(pos,(1-p[i]+M)%M,1);
    }
    printf("%lld\n",ans);
    return 0;
}