Last active
December 10, 2015 02:39
-
-
Save kaikai2/4369450 to your computer and use it in GitHub Desktop.
A丢100个6面骰子,B丢160个4面骰子。那么A的点数总和比B的点数总和大的几率以及相等的几率分别该怎么计算?
http://weibo.com/2039631434/zaMYL2jOj
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
/* | |
A丢100个6面骰子,B丢160个4面骰子。那么A的点数总和比B的点数总和大的几率以及相等的几率分别该怎么计算? | |
http://weibo.com/2039631434/zaMYL2jOj | |
*/ | |
#include <cstdio> | |
#include <cassert> | |
template<int n,int dice> | |
double f(int d) | |
{ | |
static double fn[n+1][n*dice+1] = {0}; | |
static bool inited = false; | |
if (!inited) | |
{ | |
fn[0][0] = 1; | |
for (int j = 1; j <= n; j++) | |
{ | |
for (int i = j; i <= j*dice; i++) | |
{ | |
double p = 0; | |
for (int k = 1; k <= dice; k++) | |
{ | |
if (i >= k) | |
p += fn[j-1][i - k] / (double)dice; | |
} | |
fn[j][i] = p; | |
} | |
} | |
inited = true; | |
} | |
assert(d <= n*dice+1); | |
return fn[n][d]; | |
} | |
double p_equal() | |
{ | |
double p = 0; | |
for (int d = 160; d <= 600; d++) | |
{ | |
p += f<100, 6>(d) * f<160, 4>(d); | |
} | |
return p; | |
} | |
double p_6_gt_4() | |
{ | |
double p = 0; | |
for (int a = 161; a <= 600; a++) | |
{ | |
for (int b = 160; b < a; b++) | |
{ | |
p += f<100, 6>(a) * f<160, 4>(b); | |
} | |
} | |
return p; | |
} | |
int main() | |
{ | |
printf("%lf\n%lf\n", p_equal(), p_6_gt_4()); | |
return 0; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment