## 题目描述

Given a sequence $a$ and an empty sequence $b$, each element in $a$ is added to $b$ with probability $P$ (independently to the other elements). Denote $s=\bigoplus_{i=1}^{|b|} b_i$, where $\oplus$ denotes bitwise xor and $s=0$ if $b$ is empty. Find the expected value of $s^2$.

## 算法分析

\begin{align} \sum_s s^2P(s) &= \sum_s (\sum_{i=0}^{29} [s二进制第i位为1] 2^i)^2P(s) \\ &= \sum_s P(s) \sum_{i=0}^{29} [s二进制第i位为1] 2^i \sum_{j=0}^{29} [s二进制第j位为1] 2^j \\ &= \sum_s P(s) \sum_{i=0}^{29} \sum_{j=0}^{29} [s二进制第i位和第j位都为1] 2^{i+j} \\ &= \sum_{i=0}^{29} \sum_{j=0}^{29} P(s二进制第i位和第j位都为1) 2^{i+j} \end{align}

## 代码

#include <cstdio>
#include <cstring>
#include <algorithm>

int const N = 100005, MOD = 1000000007;

int a[N], f[N];

int power(int a, int b) {
int ret = 1;
for (; b; b >>= 1) {
if (b & 1) {
ret = 1ll * ret * a % MOD;
}
a = 1ll * a * a % MOD;
}
return ret;
}

int main() {
int n, x, y;
scanf("%d%d%d", &n, &x, &y);
int p = 1ll * x * power(y, MOD - 2) % MOD, q = (MOD + 1 - p) % MOD;
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
int ans = 0, ans2 = 0;
f = 1;
for (int i = 0; i < 30; ++i) {
for (int j = i; j < 30; ++j) {
for (int k = 1; k <= n; ++k) {
for (int _i = 0; _i < 2; ++_i) {
for (int _j = 0; _j < 2; ++_j) {
int __i = _i ^ (a[k] >> i & 1);
int __j = _j ^ (a[k] >> j & 1);
f[k][_i][_j] = (1ll * f[k - 1][_i][_j] * q + 1ll * f[k - 1][__i][__j] * p) % MOD;
}
}
}
ans = (ans + (1ll << i + j + (i != j)) % MOD * f[n]) % MOD;
}
}
printf("%d\n", ans);
return 0;
}

## 题目描述

$n$ black and white balls were put into a bag. Petya doesn’t know exactly how many black balls are there among them. He knows, however, that there are $0, 1, \ldots, n$ black balls among all balls in the bag with equal probability.
Petya took $l$ balls from the bag at random, and $l_1$ of them turned out black, while $l_2$ other turned out white ($l_1+l_2=l$). Now he wants to predict how many black balls were there initially in the bag. Of course, if $l \lt n$, he can’t be sure in his prediction, but he wants to predict a segment $[a, b]$, such that the amount $k$ of black balls belongs to it with probability at least $p$.
You are given $n, l_1, l_2$ and $p$, and you must find such $a$ and $b$ that $b-a$ is minimal possible. If there are several such pairs $(a, b)$, choose the one with the smallest $a$.

## 算法分析

$$P(B_i|A)={P(B_i)P(A|B_i) \over \sum_{j=0}^n P(B_j)P(A|B_j)}$$

$$P(B_i)={1 \over n+1}, \; P(A|B_i)={{i \choose l_1}{n-i \choose l_2} \over {n \choose l}}$$

## 代码

#include <algorithm>
#include <cstdio>
#include <cstring>

static int const N = 55;
double a[N];

double get_c(int n, int m) {
if (n < m)
return 0;
double ret = 1;
for (int i = 0; i < m; ++i)
ret *= 1. * (n - i) / (m - i);
return ret;
}

int main() {
int n, l1, l2, p;
scanf("%d%d%d%d", &n, &l1, &l2, &p);
double b = 0;
for (int i = 0; i <= n; ++i)
b += a[i] = 1. / (n + 1) * get_c(i, l1) * get_c(n - i, l2) / get_c(n, l1 + l2);
for (int i = 1; i <= n + 1; ++i)
for (int j = 0; j + i - 1 <= n; ++j) {
double sum = 0;
for (int k = j; k <= j + i - 1; ++k)
sum += a[k];
if (sum / b * 100 + 1e-12 >= p)
return printf("%d %d\n", j, j + i - 1), 0;
}
return 0;
}


## 题目描述

On Panda’s Birthday party, he received a strange present from Jason. The present is a black box with $4$ dices in it which is used to play a game. The dice in the box is unusual. Instead of the digits, only red or blue is painted on each side of the dice. Before the first round of the game, the box can repaint every side of every dice red or blue with equal probability. Then for each round of the game, the box will roll the $4$ dices and tell the player the number of red side facing up, which is the point the player get. Now, Panda has play it for two rounds and he tell you the point he has got for each round. Can you tell him the expected point he can get for next round.

## 算法分析

1. 对于一个骰子，它有$t$个面被染成红色的概率$P(x=t)={{6 \choose t} \over 2^6}={{6 \choose t} \over 64}$；
2. 扔一个骰子$n$次，其中有$k$次朝上的面是红色的概率为$\sum_{i=0}^6 P(x=i) \cdot {i^k(6-i)^{n-k} \over 6^n}$。

\begin{align} P(111|11)&={9 \over 14} \\ P(101|10)&={1 \over 2} \\ P(001|00)&={5 \over 14} \end{align}

$${p+q \over 2} \cdot {9 \over 14}+\left(4-{p+q \over 2}\right) \cdot {5 \over 14}={p+q+10 \over 7}$$

$${p+q-1 \over 2} \cdot {9 \over 14}+\left(4-{p+q+1 \over 2}\right) \cdot {5 \over 14}+{1 \over 2}={p+q+10 \over 7}$$

## 题目描述

Two of the three members of the winning team of one of the ACM regional contests are going to meet in order to train for the upcoming World Finals. They decided that they will meet sometime between $X$ o’clock and $Y$ o’clock. Because they never get anywhere on time (they were late even on the day of the regional contest), they did not set an exact time when they will meet. However, they decided that the one who gets first at the meeting point will not wait more than $Z$ minutes for the other one (they calculated that, if the other one will not come within $Z$ minutes from the arrival of the first of them, then it is very probable that he will not show up at all).
Knowing that, in the end, both of them will show up at some time between $X$ o’clock and $Y$ o’clock (not necessarily after an integer number of minutes), compute which is the probability that they will actually meet.

## 代码

#include <cstdio>

using namespace std;

struct Solver {
private:
int x, y; double z;
void input() { scanf(" %d %d %lf", &x, &y, &z); }
void process() {
int all = (y - x) * 60;
double sum = 0;
if (2 * z <= all) sum = z + 2 * (all - z) * (z / all) / 2;
else sum = z + 2 * z * (1 - z / all) / 2;
printf("%.8lf\n", sum / all);
}

public:
void solve() { input(), process(); }
} solver;

int main() {
solver.solve();
return 0;
}


## 题目描述

One night, having had a hard day at work, Petya saw a nightmare. There was a binary search tree in the dream. But it was not the actual tree that scared Petya. The horrifying thing was that Petya couldn’t search for elements in this tree. Petya tried many times to choose key and look for it in the tree, and each time he arrived at a wrong place. Petya has been racking his brains for long, choosing keys many times, but the result was no better. But the moment before Petya would start to despair, he had an epiphany: every time he was looking for keys, the tree didn’t have the key, and occured exactly one mistake. “That’s not a problem!”, thought Petya. “Why not count the expectation value of an element, which is found when I search for the key”. The moment he was about to do just that, however, Petya suddenly woke up.
Thus, you are given a binary search tree, that is a tree containing some number written in the node. This number is called the node key. The number of children of every node of the tree is equal either to $0$ or to $2$. The nodes that have $0$ children are called leaves and the nodes that have $2$ children, are called inner. An inner node has the left child, that is the child whose key is less than the current node’s key, and the right child, whose key is more than the current node’s key. Also, a key of any node is strictly larger than all the keys of the left subtree of the node and strictly smaller than all the keys of the right subtree of the node.
Also you are given a set of search keys, all of which are distinct and differ from the node keys contained in the tree. For each key from the set its search in the tree is realised. The search is arranged like this: initially we are located in the tree root, if the key of the current node is larger that our search key, then we move to the left child of the node, otherwise we go to the right child of the node and the process is repeated. As it is guaranteed that the search key is not contained in the tree, the search will always finish in some leaf. The key lying in the leaf is declared the search result.
It is known for sure that during the search we make a mistake in comparing exactly once, that is we go the wrong way, but we won’t make any mistakes later. All possible mistakes are equiprobable, that is we should consider all such searches where exactly one mistake occurs. Your task is to find the expectation (the average value) of the search result for every search key, considering that exactly one mistake occurs in the search. That is, for a set of paths containing exactly one mistake in the given key search, you should count the average value of keys containing in the leaves of those paths.

## 代码

#include <cstdio>
#include <algorithm>
using namespace std;
struct node_type {
long long val, min, max, child;
} node;
long long n, k, cnt, times, root, top, id;
double ans;
void dfs(long long t) {
if (!node[t].child) {
node[t].min = node[t].max = node[t].val; return;
}
dfs(node[t].child), dfs(node[t].child);
node[t].min = node[node[t].child].min, node[t].max = node[node[t].child].max;
}
void find(long long t) {
if (!node[t].child) {
id[top] = node[t].val, ans[top++] = 1.0 * cnt / times; return;
}
cnt += node[node[t].child].min, ++times, find(node[t].child), id[top++] = node[t].val;
cnt += node[node[t].child].max - node[node[t].child].min, find(node[t].child);
cnt -= node[node[t].child].max, --times;
}
int main()
{
scanf("%lld", &n);
for (int i = 1; i <= n; ++i) {
long long a; scanf("%lld%lld", &a, &node[i].val);
if (a == -1) root = i;
else if (!node[a].child) node[a].child = i;
else {
node[a].child = i;
if (node[node[a].child].val > node[node[a].child].val)
node[a].child ^= node[a].child ^= node[a].child ^= node[a].child;
}
}
dfs(root), find(root), scanf("%lld", &k);
for (int i = top - 2; i >= 0; --i) if (!ans[i]) ans[i] = ans[i + 1];
while (k--) {
long long a, l = 0, r = top; scanf("%lld", &a);
while (l + 1 < r) {
long long mid = l + r >> 1; if (id[mid] < a) l = mid; else r = mid;
}
printf("%.10lf\n", ans[l]);
}
return 0;
}