题解 聚烷撑乙二醇

传送门

令 \(ans[i]\) 为从第 \(i\) 个位置开始向后走的最大收益
于是根据第 \(i\) 个位置随出来的数与 \(ans[i]\) 的大小关系按题意转移即可
但是卡精度，需要开long double
u1s1，long double就真的没爆吗？

• 当数据范围都是1e9级别且需要小数的时候记得开long double

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
long double l[N], r[N], ans[N];

namespace task1{
	void solve() {
		for (int i=n; i; --i) {
			if (r[i]<ans[i+1]) ans[i]=ans[i+1];
			else if (l[i]>ans[i+1]) ans[i]=(l[i]+r[i])/2;
			else if (l[i]==r[i]) ans[i]=l[i];
			else ans[i]=(r[i]-ans[i+1])/(r[i]-l[i])*(r[i]+ans[i+1])/2 + (ans[i+1]-l[i])/(r[i]-l[i])*ans[i+1];
		}
		printf("%.5Lf\n", ans[1]);
		exit(0);
	}
}

signed main()
{
	freopen("pag.in", "r", stdin);
	freopen("pag.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) l[i]=read(), r[i]=read();
	task1::solve();

	return 0;
}