HDU 5245 Joyful (期望)

Posted 2020-09-15 dwtfukgv

题意：进行K次染色，每次染色会随机选取一个以(x1,y1),(x2,y2)为一组对角的子矩阵进行染色，求K次染色后染色面积的期望值(四舍五入)。

析：我们可以先求出每个格子的期望，然后再加起来即可。我们可以把格子进行划分，然后再求概率。

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>
#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 freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "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 = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -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;
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    int k;
    cin >> n >> m >> k;
    double ans = 0.0;
    for(int i = 1; i <= n; ++i)
      for(int j = 1; j <= m; ++j){
        double p = m * n;
        p += 1.0 * (i-1) * (j-1) * (n-i+1) * (m-j+1);
        p += 1.0 * (i-1) * (m-j) * (n-i+1) * j;
        p += 1.0 * (j-1) * (n-i) * (m-j+1) * i;
        p += 1.0 * (n-i) * (m-j) * i * j;
        p += 1.0 * (i-1) * m * (n-i+1);
        p += 1.0 * (m-j) * n * j;
        p += 1.0 * (n-i) * m * i;
        p += 1.0 * (j-1) * n * (m-j+1);
        p = p / (1.0*n*n*m*m);
        ans += 1.0 - (pow(1.0-p, k));
      }
    printf("Case #%d: %.f\n", kase, ans);
  }
  return 0;
}