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  • POJ 3440 Coin Toss(概率)

    题目链接

    概率问题,像是概率论上学的均匀分布,是不是呢,忘了。。。

    概率同面积有关系,我写的各种搓,然后此题格式十分变态,=前有的时候俩空格,有的时候一个空格。代码各种搓。

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 #include <cmath>
     5 #include <algorithm>
     6 using namespace std;
     7 #define PI 3.141592653
     8 #define eps 1e-9
     9 int main()
    10 {
    11     int t,cas = 1;
    12     double n,m,d,r,S;
    13     double t11,t12,t13,t14,t21,t22,t23,t24,t31,t32,t33,t34;
    14     double a1,a2,a3,a4;
    15     scanf("%d",&t);
    16     while(t--)
    17     {
    18         scanf("%lf%lf%lf%lf",&n,&m,&d,&r);
    19         S = n*m*d*d;
    20         t11 = (d-r/2)*(d-r/2);
    21         t12 = 2*(d-r/2)*r/2;
    22         t13 = 0.25*PI*(r/2)*(r/2);
    23         t14 = (r/2)*(r/2) - t13;
    24         t21 = (d-r/2)*(d-r);
    25         t22 = (d-r/2)*r + (d-r)*r/2;
    26         t23 = 0.5*PI*(r/2)*(r/2);
    27         t24 = 2*(r/2)*(r/2) - t23;
    28         t31 = (d-r)*(d-r);
    29         t32 = 4*(d-r)*r/2;
    30         t33 = PI*(r/2)*(r/2);
    31         t34 = r*r - t33;
    32         if(n == 1&&m == 1)
    33         {
    34             a1 = S;
    35             a2 = 0;
    36             a3 = 0;
    37             a4 = 0;
    38         }
    39         else if(n == 1)
    40         {
    41             a1 = 2*(d-r/2)*d + (m-2)*(d-r)*d;
    42             a2 = S - a1;
    43             a3 = 0;
    44             a4 = 0;
    45         }
    46         else if(m == 1)
    47         {
    48             a1 = 2*(d-r/2)*d + (n-2)*(d-r)*d;
    49             a2 = S - a1;
    50             a3 = 0;
    51             a4 = 0;
    52         }
    53         else
    54         {
    55             a1 = 4*t11 + 2*(n-2+m-2)*t21 + (n-2)*(m-2)*t31;
    56             a2 = 4*t12 + 2*(n-2+m-2)*t22 + (n-2)*(m-2)*t32;
    57             a3 = 4*t13 + 2*(n-2+m-2)*t23 + (n-2)*(m-2)*t33;
    58             a4 = 4*t14 + 2*(n-2+m-2)*t24 + (n-2)*(m-2)*t34;
    59         }
    60         printf("Case %d:
    ",cas++);
    61         printf("Probability of covering 1 tile  = %.4f%%
    ",a1*100/S);
    62         printf("Probability of covering 2 tiles = %.4f%%
    ",a2*100/S);
    63         printf("Probability of covering 3 tiles = %.4f%%
    ",a4*100/S);
    64         printf("Probability of covering 4 tiles = %.4f%%
    ",a3*100/S);
    65         printf("
    ");
    66     }
    67     return 0;
    68 }
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  • 原文地址:https://www.cnblogs.com/naix-x/p/3207548.html
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