题目描述
BuggyD loves to carry his favorite die around. Perhaps you wonder why it's his favorite? Well, his die is magical and can be transformed into an N-sided unbiased die with the push of a button. Now BuggyD wants to learn more about his die, so he raises a question:
What is the expected number of throws of his die while it has N sides so that each number is rolled at least once?
输入格式
The first line of the input contains an integer t, the number of test cases. t test cases follow.
Each test case consists of a single line containing a single integer N (1 <= N <= 1000) - the number of sides on BuggyD's die.
输出格式
For each test case, print one line containing the expected number of times BuggyD needs to throw his N-sided die so that each number appears at least once. The expected number must be accurate to 2 decimal digits.
题意翻译
一个n面的骰子，求期望掷几次能使得每一面都被掷到。
输入格式
2
1
12
输出格式
1.00
37.24
思路
令\(dp_i\)为表示当前已经得到了 ii 个面后的期望次数。那么我们会有\(2\)种情况
\(1:\)掷出已经出现过的面，概率为\(\frac{i}{n}\)
\(2:\)掷出没有出现过的面，概率为\(\frac{n-i}{n}\)
则有：
\(dp_i=\frac{i}{n}*dp_i+\frac{n-i}{n}*dp_{i+1}+1\)
\(\frac{n-i}{n}*dp_i=\frac{n-i}{n}*dp_{i+1}+1\)
\(dp_i=dp_{i+1}+\frac{n}{n-i}\)
边界条件为\(dp_n=0\)，倒推即可求出答案。
代码
/************************************************
*Author : xzj213
*Created Time : 2020.01.30.17:44
*Mail : [email protected]
*Problem : SP1026
************************************************/
#include
#define REP(i,a,b) for(register int i=(a);i<=(b);i++)
#define DREP(i,a,b) for(register int i=(a);i>=(b);i--)
#define mem(a,x) memset((a),(x),sizeof(a))
#define pii pair
#define lson k<<1
#define rson k<<1|1
#define x first
#define y second
#define str(a) strlen(a)
using namespace std;
const int maxn=1000+5;
int n,T;
double dp[maxn];
void chkmax(int &a,int b){if(ab)a=b;}
int read() {
int x=0,f=1;
char ch=getchar();
while(ch>57 || ch<48){if(ch==45)f=-1;ch=getchar();}
while(ch<=57 && ch>=48){x=x*10+ch-48;ch=getchar();}
return x*f;
}
int main() {
freopen("SP1026.in","r",stdin);
freopen("SP1026.out","w",stdout);
T=read();
while(T--) {
n=read();
dp[n]=0;
DREP (i,n-1,0) dp[i]=dp[i+1]+1.0*n/(n-i);
printf("%.2lf\n",dp[0]);
}
return 0;
}