CF641D. Little Artem and Random Variable

Solution

設給定的兩個序列為$m{x}_{1..n},m{n}_{1..n}$。
令第一個骰子投到$1..n$的概率為$p_{1..n}$
令第二個骰子投到$1..n$的概率為$q_{1..n}$
顯然有
$$m{x}_i=(\sum _{j\le i}{p}_j)(\sum _{j\le i}{p}_j)?(\sum _{j<i}{p}_j)(\sum _{j<i}{p}_j)$$
$$m{n}_i=(\sum _{j\ge i}{p}_j)(\sum _{j\ge i}{p}_j)?(\sum _{j>i}{p}_j)(\sum _{j>i}{p}_j)$$
可以把${\sum }_{j\le x}{p}_j$當成橫坐標，把${\sum }_{j\le y}{q}_j$當成縱坐標，這樣就形成一個$1?1$的矩形，$m{x}_i.m{n}_i$分別為其中一個$L$形矩陣面積，可以求得：
$$1?\sum _{j\le i}m{x}_i+m{n}_{i+1}=(\sum _{j>i}{p}_j)+(\sum _{j>i}{q}_j)$$
$$\sum m{n}_{i+1}=(\sum _{j>i}{p}_j)(\sum _{j>i}{q}_j)$$
這樣就可以通過求解一元二次方程求出$({\sum }_{j>i}{p}_j)$和$({\sum }_{j>i}{q}_j)$，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; }

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