Test instructions: Given N-class bugs, and s subsystems, you can find a bug every day, find out the N type of bug, and s have at least one of the expectations of how much.
Analysis: Should be a very simple probability dp,dp[i][j] indicates that I have been from the J subsystem to find the type I bug, to achieve the desired number of days of the target,
Obviously dp[n][s] = 0.0, and dp[0][0] is the answer, the rest is relatively simple,
DP[I][J] = (dp[i+1][j]* (n-i) *j + dp[i][j+1]*i* (s-j) + dp[i+1][j+1]* (n-i) * (s-j) + n*s)/(n*s*1.0-i*j*1.0);
The code is as follows:
#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> #define Lson l,m,rt<<1#define Rson m+1,r,rt<<1|1//#include <tr1/ unordered_map> #define FREOPENR freopen ("In.txt", "R", stdin) #define FREOPENW freopen ("OUT.txt", "w", stdout) using Namespace std;//using namespace std:: tr1;typedef Long Long ll;typedef pair<int, int> p;const int INF = 0x3f3f3f3f; Const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const Double PI = ACOs ( -1.0); const double EPS = 1e-8;const int MAXN = 1e3 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1,-1, -1};const int dc[] = {0, 1, 0,-1, 1,-1, 1, -1};const char *hex[] = {"0000 "," 0001 "," 0010 "," 0011 "," 0100 "," 0101 "," 0110 "," 0111 "," 1000 "," 1001 "," 1010 "," 1011 "," 1100 "," 1101 "," 1110 "," 1111 "};inline ll GCD (ll A, ll b) {return b = = 0? A:GCD (b, a%b); }int N, m;const int mon[] = {0, 31, 29, 31, 30, 31, 0, +, +,, +, +, +, +, +, 31};const int monn[] N, H, H, C, h, 31};inline int Min (int a, int b) {return a < b? A:b;} inline int Max (int a, int b) {return a > b a:b;} inline ll Min (ll A, ll b) {return a < b a:b;} inline ll Max (ll A, ll b) {return a > b a:b;} inline bool Is_in (int r, int c) {return R >= 0 && r < n && C >= 0 && C < m;} Double Dp[maxn][maxn];int Main () {int s; while (scanf ("%d%d", &n, &s) = = 2) {Dp[n][s] = Dp[n+1][s] = dp[n][s+1] = dp[n+1][s+1] = 0.0; for (int i = n; i >= 0; i.) for (int j = s; J >= 0;--j) {if (i = = N && j = = s) con Tinue; DP[I][J] = (dp[i+1][j]* (N-i) *j + dp[i][j+1]*i* (s-j) + dp[i+1][j+1]* (n-i) * (s-j) + n*s)/(n*s*1.0-i*j*1.0); } printf ("%.4f\n", dp[0][0]); } return 0;}
POJ 2096 Collecting Bugs (probabilistic DP)