HDU 2909 - Expected Allowance
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计算期望值的一道题,题目大意:给出一个n个骰子,每个骰子有m个面,给出一个削减值k,也就是所有的骰子点数得和要减去k,就是他能得到的钱,题目又给出:如果骰子的点数小于k,则得到的钱也至少是1;要求出他能得到钱的期望值;所有点数的概率是类似于一个二项式的,然后分别乘以其权值,然后除以m^n;就是所要求期望值
AC code:
#include<cstdio>
#include<cstring>
using namespace std;
int num[2][100000];
int main()
{
int n,m,k;
while(scanf(\"%d%d%d\",&n,&m,&k)!=EOF&&(n||m||k))
{
memset(num,0,sizeof(num));
for(int i=1; i<=m; i++)
num[1][i]=1;
for(int i=2; i<=n; i++)
{
for(int j=1; j<=i*m; j++)
for(int l=1; l<=m; l++)
num[i%2][j+l]+=num[(i-1)%2][j];
memset(num[(i-1)%2],0,sizeof(num[(i-1)%2]));
}
double res=0.0;
int i,j;
for(i=1; i<=k+1; i++)
res+=num[n%2][i];
for(j=2;i<=n*m; i++,j++)
res+=num[n%2][i]*j;
int count=1;
for(int i=1;i<=n;i++)
count*=m;
res/=count;
printf(\"%.8lf\\n\",res);
}
return 0;
}
AC code:
#include<cstdio>
#include<cstring>
using namespace std;
int num[2][100000];
int main()
{
int n,m,k;
while(scanf(\"%d%d%d\",&n,&m,&k)!=EOF&&(n||m||k))
{
memset(num,0,sizeof(num));
for(int i=1; i<=m; i++)
num[1][i]=1;
for(int i=2; i<=n; i++)
{
for(int j=1; j<=i*m; j++)
for(int l=1; l<=m; l++)
num[i%2][j+l]+=num[(i-1)%2][j];
memset(num[(i-1)%2],0,sizeof(num[(i-1)%2]));
}
double res=0.0;
int i,j;
for(i=1; i<=k+1; i++)
res+=num[n%2][i];
for(j=2;i<=n*m; i++,j++)
res+=num[n%2][i]*j;
int count=1;
for(int i=1;i<=n;i++)
count*=m;
res/=count;
printf(\"%.8lf\\n\",res);
}
return 0;
}
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