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POJ 3440 Coin Toss

高中概率的几何概型,这也叫作题,不过输出真的很坑。


题目大意:

n*m个边长为t的正方形组成的矩形。往矩形上抛一个直径为c的硬币,问覆盖1,2,3,4个矩形的概率为多少?


解题思路:

计算出覆盖1,2,3,4个矩形时硬币圆心可以在的位置区域。就能求出概率了~



下面是代码:

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define eps 1e-6
#define pi acos(-1.0)
#define inf 107374182
#define inf64 1152921504606846976
#define lc l,m,tr<<1
#define rc m + 1,r,tr<<1|1
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE))
#define clearall(A, X) memset(A, X, sizeof(A))
#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))
#define memcopyall(A, X) memcpy(A , X ,sizeof(X))
#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )
#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )

using namespace std;

int main()
{
    int T,case1=1;
    double n,m,t,c,ans[4];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf%lf",&n,&m,&t,&c);
        ans[0]=(t-c/2)*(t-c/2)*4 + (t-c)*(t-c/2)*(2*m+2*n-8) + (t-c)*(t-c)*(n-2)*(m-2);  //覆盖一个方格时硬币圆心可以在的位置
        ans[2] = (c*c - pi*(c/2)*(c/2) )*(n-1)*(m-1);
        ans[3] = pi*(c/2)*(c/2)*(n-1)*(m-1);
        ans[1] = n*m*t*t - ans[0] - ans[2] - ans[3];
        printf("Case %d:\n",case1++);
        printf("Probability of covering 1 tile  = %.4f%%\n",ans[0]*100.0/(n*m*t*t));
        for(int i=1;i<4;i++)
        {
            printf("Probability of covering %d tiles = %.4f%%\n",i+1,ans[i]*100.0/(n*m*t*t));
        }
        puts("");
    }
    return 0;
}