ACdream 1113 The Arrow （概率DP）

ACdream 1113 The Arrow （概率DP）：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=114332#problem/E 传送门：nefu

题面：

Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu

Submit Status

Description

The history shows that We need heroes in every dynasty. For example, Liangshan Heroes. People hope that these heroes can punish the bad  guys and recover the justice. Nowadays, we also need heroes to provent us from being chopped, or being attacked by a bomb. 

Kuangbin is a very very very very very.... (very * 1e9 ) good boy, after he knows The Arrow, he want to be The Arrow of China. But he is also a little afraid of being killed by the bad guys. So he decides to throw dices to make the decision.

The dice is a cube with 1 2 3 4 5 6 on it's sides. When he throws a dice, every number is of the same probablity to appear. He will write down a number N in the paper at first, and then throw the dice. When the sum of the number he throwed is less than N, he will keep throwing. But if the sum exceeds N, this throwing does not count.

For example, If the sum is 5,and N is 6, if we throw 2, 5 + 2 > 6, then  the sum keeps to be 5.

If he can end the throwing in a certain time, he will make the decision to become The Arrow.

Now , kuangbin wonders that what's the expectation of the time of throwing dices.

Input

First line, The number of cases t <= 100

In the next t lines, there will be t numbers.

every number is not bigger than 100000

Output

Each test output a number rounded to the second digit.

Sample Input

11

Sample Output

6.00

题目大意：

一边掷骰子一边求点数之和，如果当前点数之和为sum，再掷一次骰子之后得到点数为y，且满足sum+y>n（n为给定值）时，保持sum不变，求所掷骰子的次数的期望。

题目分析：

dp公式为：

dp[i]=dp[i]*y/6+dp[i+1]*y/6+dp[i+2]*y/6+...+dp[i+6]*y/6+1.

代码实现：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;double dp[100010];int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(int i=n-1;i>=0;i--)        {            int tot=0;            double temp=0;            for(int j=1;j<=6;j++)            {                if(i+j>n) tot++;                else temp+=dp[i+j]/6.0;            }            dp[i]=(temp+1)/(6-tot)*6;        }        printf("%.2lf\n",dp[0]);    }    return 0;}