牛客网暑期ACM多校训练营（第五场）F：take（树状数组）

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

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K 64bit IO Format: %lld
题目描述
Kanade has $n$ boxes , the i-th box has $p[i]$ probability to have an diamond of $d[i]$ size.

At the beginning , Kanade has a diamond of 0 size. She will open the boxes from 1-st to n-th. When she open a box,if there is a diamond in it and it’s bigger than the diamond of her , she will replace it with her diamond.

Now you need to calculate the expect number of replacements.

You only need to output the answer module $998244353$.

Notice: If $x$%$998244353=y*d$%$998244353$ ,then we denote that
$x/y$%$998244353 =d$%$998244353$

输入描述:
The first line has one integer $n$.

Then there are $n$ lines. each line has two integers $p[i]*100$ and $d[i]$.
输出描述:
Output the answer module 998244353
示例1
3
50 1
50 2
50 3
输出
499122178
备注:
$1<= n <= 100000$
$1<=p[i]*100 <=100$
$1<=d[i]<=10^9$

思路：分别考虑每个箱子钻石的贡献。
对于某个箱子，若要和这个箱子里的钻石交换，那么位于这个箱子之前的且比当前钻石大的钻石都不能被拿到；而比当前钻石小的，拿到拿不到没有影响；在它后面的钻石对这次交换也没影响。这样你就必定会与当前的钻石交换。
于是我们可以求出能交换到$d[i]$这个钻石的概率$Pi$，那么这个箱子对答案的贡献是$1*Pi$。
所以$ans=\sum_{i=1}^{n}P_i$
可以用树状数组维护更新

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e5+10;
const int MOD=998244353;
const long long INF=1e18;
typedef long long ll;
struct lenka
{
    int p,w,id;
}a[MAX];
int cmp(const lenka &A,const lenka& B)
{
    if(A.w==B.w)return A.id<B.id;
    return A.w>B.w;
}
ll A[MAX];
void add(int x,int n,int y){while(x<=n){(A[x]*=y)%=MOD;x+=x&(-x);}}
ll ask(int x){ll ans=1;while(x){(ans*=A[x])%=MOD;x-=x&(-x);}return ans;}
int main()
{
    int inv=828542813;//100的逆元
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].p,&a[i].w);
        a[i].id=i;
    }
    sort(a+1,a+n+1,cmp);       //按权值从大到小排序
    for(int i=1;i<=n;i++)A[i]=1;
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        ans+=ask(a[i].id-1)*a[i].p%MOD*inv%MOD;//查询在它之前比它大的概率乘积
        ans%=MOD;
        add(a[i].id,n,1ll*(100-a[i].p)*inv%MOD);
    }
    cout<<ans<<endl;
    return 0;
}