CF641D. Little Artem and Random Variable

Solution

设给定的两个序列为 $mx_{1..n},mn_{1..n}$ 。
令第一个骰子投到 $1 . . n$ 的概率为 $p_{1..n}$
令第二个骰子投到 $1 . . n$ 的概率为 $q_{1..n}$
显然有
$mxi=(∑j≤ipj)(∑j≤ipj)−(∑j<ipj)(∑j<ipj)mx_i=(\sum_{j\leq i}p_j)(\sum_{j\leq i}p_j)-(\sum_{j<i}p_j)(\sum_{j<i}p_j)$
$mni=(∑j≥ipj)(∑j≥ipj)−(∑j>ipj)(∑j>ipj)mn_i=(\sum_{j\geq i}p_j)(\sum_{j\geq i}p_j)-(\sum_{j>i}p_j)(\sum_{j>i}p_j)$
可以把 $∑j≤xpj\sum_{j\leq x}p_j$ 当成横坐标，把 $∑j≤yqj\sum_{j\leq y}q_j$ 当成纵坐标，这样就形成一个 $1 * 1$ 的矩形， $mx_i.mn_i$ 分别为其中一个 $L$ 形矩阵面积，可以求得：
$1−∑j≤imxi+mni+1=(∑j>ipj)+(∑j>iqj)1-\sum_{j\leq i}mx_i+mn_{i+1}=(\sum_{j>i}p_j)+(\sum_{j>i}q_j)$
$∑mni+1=(∑j>ipj)(∑j>iqj)\sum mn_{i+1}=(\sum_{j>i}p_j)(\sum_{j>i}q_j)$
这样就可以通过求解一元二次方程求出 $(∑j>ipj)(\sum_{j>i}p_j)$ 和 $(∑j>iqj)(\sum_{j>i}q_j)$ ，差分即可。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second
#define int llusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-15;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=100005;
const int INF=0x7fffffff;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
double mx[MAXN],mn[MAXN],smx[MAXN],smn[MAXN],sp[MAXN],sq[MAXN];
signed main()
{int n=read();for (int i=1;i<=n;i++) scanf("%lf",&mx[i]);for (int i=1;i<=n;i++) scanf("%lf",&mn[i]);for (int i=1;i<=n;i++) smx[i]=smx[i-1]+mx[i];for (int i=n;i>=1;i--) smn[i]=smn[i+1]+mn[i];for (int i=1;i<=n;i++){double pl=1+smx[i]-smn[i+1],mul=smx[i],mi=sqrt(pl*pl-mul*4+eps);sp[i]=(pl+mi)*0.5,sq[i]=(pl-mi)*0.5;}for (int i=1;i<=n;i++) printf("%.10lf ",sp[i]-sp[i-1]); puts("");for (int i=1;i<=n;i++) printf("%.10lf ",sq[i]-sq[i-1]); puts("");return 0;
}