牛客网暑期ACM多校训练营(第五场): F. take（期望+线段树）

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

输入

输出

499122178

算一下每个箱子对答案的贡献，加在一起就可以了

对于第i个箱子，它对答案的贡献就为

也就是当且仅当当前箱子里出现钻石，且前面所有钻石比它大的箱子都不出现钻石，在经过这个箱子时才会replace（对答案有贡献）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
#define LL long long
#define mod 998244353
LL tre[500005];
typedef struct Res
{int x, id, p;bool operator < (const Res &b) const{if(x>b.x || x==b.x && id<b.id)return 1;return 0;}
}Res;
Res s[200005];
void Create(int l, int r, int x)
{int m;if(l==r){tre[x] = 1;return;}m = (l+r)/2;Create(l, m, x*2);Create(m+1, r, x*2+1);tre[x] = 1;
}
void Update(int l, int r, int x, int a, int b)
{int m;if(l==r){tre[x] = (LL)(100-b)*828542813%mod;return;}m = (l+r)/2;if(a<=m)Update(l, m, x*2, a, b);elseUpdate(m+1, r, x*2+1, a, b);tre[x] = tre[x*2]*tre[x*2+1]%mod;
}
LL Query(int l, int r, int x, int a, int b)
{int m;LL ans = 1;if(l>=a && r<=b)return tre[x];m = (l+r)/2;if(a<=m)ans = ans*Query(l, m, x*2, a, b)%mod;if(b>=m+1)ans = ans*Query(m+1, r, x*2+1, a, b)%mod;return ans;
}
LL Pow(LL a, LL b)
{LL ans = 1;while(b){if(b%2)ans = ans*a%mod;a = a*a%mod;b /= 2;}return ans;
}
int main(void)
{LL ans;int n, i;scanf("%d", &n);for(i=1;i<=n;i++){scanf("%d%d", &s[i].p, &s[i].x);s[i].id = i;}sort(s+1, s+n+1);Create(1, n, 1);ans = 0;for(i=1;i<=n;i++){ans = (ans+Query(1, n, 1, 1, s[i].id)*s[i].p%mod*828542813)%mod;Update(1, n, 1, s[i].id, s[i].p);}printf("%lld\n", ans);return 0;
}