atcoderI - Coins ( 概率DP)
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I - Coins
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 100100 points
Problem Statement
Let NN be a positive odd number.
There are NN coins, numbered 1,2,…,N1,2,…,N. For each ii (1≤i≤N1≤i≤N), when Coin ii is tossed, it comes up heads with probability pipi and tails with probability 1−pi1−pi.
Taro has tossed all the NN coins. Find the probability of having more heads than tails.
Constraints
- NN is an odd number.
- 1≤N≤29991≤N≤2999
- pipi is a real number and has two decimal places.
- 0<pi<10<pi<1
Input
Input is given from Standard Input in the following format:
NN p1p1 p2p2 …… pNpN
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10−910−9.
Sample Input 1 Copy
3 0.30 0.60 0.80
Sample Output 1 Copy
0.612
The probability of each case where we have more heads than tails is as follows:
- The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Head)(Coin1,Coin2,Coin3)=(Head,Head,Head) is 0.3×0.6×0.8=0.1440.3×0.6×0.8=0.144;
- The probability of having (Coin1,Coin2,Coin3)=(Tail,Head,Head)(Coin1,Coin2,Coin3)=(Tail,Head,Head) is 0.7×0.6×0.8=0.3360.7×0.6×0.8=0.336;
- The probability of having (Coin1,Coin2,Coin3)=(Head,Tail,Head)(Coin1,Coin2,Coin3)=(Head,Tail,Head) is 0.3×0.4×0.8=0.0960.3×0.4×0.8=0.096;
- The probability of having (Coin1,Coin2,Coin3)=(Head,Head,Tail)(Coin1,Coin2,Coin3)=(Head,Head,Tail) is 0.3×0.6×0.2=0.0360.3×0.6×0.2=0.036.
Thus, the probability of having more heads than tails is 0.144+0.336+0.096+0.036=0.6120.144+0.336+0.096+0.036=0.612.
Sample Input 2 Copy
1 0.50
Sample Output 2 Copy
0.5
Outputs such as 0.500
, 0.500000001
and 0.499999999
are also considered correct.
Sample Input 3 Copy
5 0.42 0.01 0.42 0.99 0.42
Sample Output 3 Copy
0.3821815872
double p[maxn];
double dp[3050][3050];
int n;
题意:给N个硬币,每一个硬币扔向空中落地是正面朝上的概率是p[i] ,让求扔了N个硬币,正面的数量大于背面数量的概率。
很裸的概率DP,我们思考一下状态和转移方程。
我们这样定义状态,定义dp[i][j] 为到第i个硬币时有j个是正面的概率。那么所求答案为sum{ dp[n][i] || (n+1)/2<=i<=n}
题目说了n为odd,
那么状态转移即为: dp[i][j]=dp[i-1][j-1]*p[i]+dp[i-1][j]*(1.0-p[i]);
意思为,到了第i个硬币时,j个正面朝上的状态可以由以下两个状态转移过来:
1、第i-1个硬币的时候,有j-1个正面朝上的,第i个硬币也正面朝上。
2、第i-1个硬币的时候,有j个正面朝上的,第i个硬币反面朝上。
然后初始状态定义
dp[1][1]=p[1];
dp[1][0]=1.0000000-p[1];
注意处理下边界情况就好了。
细节见AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘ ‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) using namespace std; typedef long long ll; inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ double p[maxn]; double dp[3050][3050]; int n; int main() { gg(n); repd(i,1,n) { scanf("%lf",&p[i]); } dp[1][1]=p[1]; dp[1][0]=1.0000000-p[1]; repd(i,2,n) { for(int j=0;j<=i;j++) { if(j==0) { dp[i][j]=dp[i-1][j]*(1.00000-p[i]); continue; } dp[i][j]=dp[i-1][j-1]*p[i]+dp[i-1][j]*(1.0000-p[i]); // dp[i][j-1]=dp[i-1][j-1]*(1.000000-p[i]); } } double ans=0.0000000000000000000; for(int i=(n+1)/2;i<=n;i++) { ans+=dp[n][i]; } printf("%.10lf ", ans); return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘ ‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
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