Candy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1159 Accepted Submission(s): 508
Special Judge
Problem Description
LazyChild is a lazy child who likes candy very much. Despite being very young, he has two large candy boxes, each contains n candies initially. Everyday he chooses one box and open it. He chooses the first box with probability p and the second box with probability (1 - p). For the chosen box, if there are still candies in it, he eats one of them; otherwise, he will be sad and then open the other box.
He has been eating one candy a day for several days. But one day, when opening a box, he finds no candy left. Before opening the other box, he wants to know the expected number of candies left in the other box. Can you help him?
He has been eating one candy a day for several days. But one day, when opening a box, he finds no candy left. Before opening the other box, he wants to know the expected number of candies left in the other box. Can you help him?
Input
There are several test cases.
For each test case, there is a single line containing an integer n (1 ≤ n ≤ 2 × 105) and a real number p (0 ≤ p ≤ 1, with 6 digits after the decimal).
Input is terminated by EOF.
For each test case, there is a single line containing an integer n (1 ≤ n ≤ 2 × 105) and a real number p (0 ≤ p ≤ 1, with 6 digits after the decimal).
Input is terminated by EOF.
Output
For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is a real number indicating the desired answer.
Any answer with an absolute error less than or equal to 10-4 would be accepted.
Any answer with an absolute error less than or equal to 10-4 would be accepted.
Sample Input
10 0.400000
100 0.500000
124 0.432650
325 0.325100
532 0.487520
2276 0.720000
Sample Output
Case 1: 3.528175
Case 2: 10.326044
Case 3: 28.861945
Case 4: 167.965476
Case 5: 32.601816
Case 6: 1390.500000
Source
Recommend
liuyiding
公式很好推。。。
主要是组合数的计算
/* * Author: * Created Time: 2013/10/16 18:29:51 * File Name: A.cpp * solve: A.cpp */ #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<string> #include<map> #include<stack> #include<set> #include<iostream> #include<vector> #include<queue> //ios_base::sync_with_stdio(false); //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define sz(v) ((int)(v).size()) #define rep(i, a, b) for (int i = (a); i < (b); ++i) #define repf(i, a, b) for (int i = (a); i <= (b); ++i) #define repd(i, a, b) for (int i = (a); i >= (b); --i) #define clr(x) memset(x,0,sizeof(x)) #define clrs( x , y ) memset(x,y,sizeof(x)) #define out(x) printf(#x" %d ", x) #define sqr(x) ((x) * (x)) typedef long long LL; const int INF = 1000000000; const double eps = 1e-8; const int maxn = 210000; int sgn(const double &x) { return (x > eps) - (x < -eps); } double lnx[maxn*2]; void init() { lnx[0] = log(1.0); lnx[1] = log(1.0); repf(i,2,400010) { lnx[i] = lnx[i - 1] + log((double)i); } } double C(int n,int m) { return lnx[n] - lnx[m] - lnx[n - m]; } int main() { //freopen("in.txt","r",stdin); int n; double p; int k = 1; init(); while(scanf("%d%lf",&n,&p) == 2) { double ans = 0; double P = log(p); double Q = log(1 - p); rep(i,0,n) { double tmp = C(n+i,i); ans += (n-i)*(exp(tmp + (n+1)*P + i*Q) + exp(tmp + (n+1)*Q + i*P)); } printf("Case %d: %.6lf ",k++,ans); } return 0; }