【題意】:有a,b,c三種物品,a和b有無限多件,c只有s件,現在Amjad想拿c這個物品,但是他前面有n個人。現在每人輪流拿一個物品,前n個人是隨機拿,而Amjad是有目的性去拿,問Amjad取得c的概率是多少。
【題解】:概率dp。
設dp[i][j]表示前 i 個人取完還有 j 個 C的概率。
轉移 :dp[i][j] = dp[i-1][j] * 2 / 3 + dp[i-1][j+1] / 3;
最后ans = ( ∑dp[n][i], 1 <= i <= s ) / ( ∑dp[n][i], 0 <= i <= s)。
實現時使用了滾動數組。
【代碼】:
1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 double dp[60];
19 int n, s;
20 double sum;
21 int main() {
22 while(~scanf("%d%d", &n, &s)) {
23 memset(dp, 0, sizeof(dp));
24 dp[s] = 1.0, sum = 0.0;
25 for(int i = 1; i <= n; i++)
26 for(int j = 0; j <= s; j++)
27 dp[j] = dp[j] * 2 / 3 + dp[j+1] / 3;
28 for(int i = 0; i <= s; i++) sum += dp[i];
29 double ans = 100.0 * (sum - dp[0]) / sum;
30 printf("%.5f\n", ans);
31 }
32 return 0;
33 }
34