UVa 11427 Expect the Expected (数学期望 + 概率DP)

时间:2022-12-16 06:17:24

题意:某个人每天晚上都玩游戏,如果第一次就䊨了就高兴的去睡觉了,否则就继续直到赢的局数的比例严格大于 p,并且他每局获胜的概率也是 p,但是你最玩 n 局,但是如果比例一直超不过 p 的话,你将不高兴的去睡觉,并且以后再也不玩了,现在问你,平均情况下他玩几个晚上游戏。

析:先假设第一天晚上就不高兴的去睡觉的概率是 q,那么有期望公式可以得到 E = q + (1-q) * (E + 1),其中 E 就是数学期望,那么可以解得 E = 1/ q,所以答案就是 1 / q,这个公式是什么意思呢,把数学期望分成两类,第一类就是第一天晚上就不再玩了,概率是 q,期望就是 1,第二类就是第一天高兴的睡觉,概率就是 1 - q,期望就是 E + 1。现在问题就是怎么求 q,这就是一个概率DP,dp[i][j] 表示玩 i 局,胜了 j 局的概率,并且要保证,胜的比例不超过 p,这样最后把所有的概率加起来就是数学期望,转移方程是 dp[i][j] = dp[i-1][j] * (1-p) + dp[i-1][j-1] * p。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int maxm = 100 + 2;
const LL mod = 100000000;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}

double dp[maxn][maxn];

int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
int molecule, denominator;
scanf("%d/%d %d", &molecule, &denominator, &n);
double p = molecule * 1. / denominator;
ms(dp, 0); dp[0][0] = 1.;
for(int i = 1; i <= n; ++i){
dp[i][0] = dp[i-1][0] * (1-p);
for(int j = 1; denominator * j <= molecule * i; ++j)
dp[i][j] += dp[i-1][j-1] * p + dp[i-1][j] * (1-p);
}
double ans = dp[n][0];
for(int j = 1; denominator * j <= molecule * n; ++j)
ans += dp[n][j];
printf("Case #%d: %d\n", kase, (int)(1. / ans));
}
return 0;
}