dice

我太傻逼了打了13.5K 思路很简单,代码很难 #include<bits/stdc++.h> using namespace std; #define ll long long struct moo{ ll opt1;//第一个 ll xiaoyu,xiaoyudengyu,dayu,dayudengyu; ll opt2;//第二个 ll val1;//第一个数 ll val2;//第二个数 ll wenhao;//有无问号 }node[110]; ll n; ll mit[10],spd[10],san[10],kno[10]; double f[110][10][10][10][10],ans1[110],ans2[110],ans3[110],ans4[110],dice[30][75]; ll needpan(ll x){ if((!node[x].xiaoyu)&&(!node[x].dayu)&&(!node[x].xiaoyudengyu)&&(!node[x].dayudengyu)) return 0; return 1; } char ch1[110],ch2[110],ch3[110],ch4[110]; ll work(char c){ if(c=='i') return 1; if(c=='p') return 2; if(c=='a') return

My function called RollDice simulates the rolling of a given number of six sided dice a given number of times. The function has two input arguments, the number of dice (NumDice) that will be rolled in each experiment and the total number (NumRolls) of times that the dice will be rolled. The output of the function will be a vector SumDice of length NumRolls that contains the sum of the dice values in each experiment. This is my code right now: how do I account for the SUM of the dice? Thanks! function SumDice= RollDice(NumDice,NumRolls) FACES= 6; maxOut= FACES*NumDice; count= zeros(1,maxOut);

Hi guys I need help with this question: A dice rolling game is played with two six-sided dice. A user playing the game, will roll the two dice and two random numbers between one and six will be generated. The sum of the two numbers will be used to decide the next step. If the sum is 2,3 or 12 then the player wins. If the sum is 7 or 11 then he/she loses. If the sum is 4, 5, 6, 8, 9 or 10 then the program automatically rolls the dice again until the player wins or loses. After every dice roll, the player will be prompted for an input. The player should decide if the game should continue or not.

题目链接： LightOJ - 1248 Description Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal. For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 0.5, and the chance of getting the same side is 0.5. So, the result is \(1 + (1 + 0.5 * (1 + 0.5 * ...))\) \(= 2

Dice Similarity Coefficent vs. IoU Several readers emailed regarding the segmentation performance of the FCN-8s model I trained in Chapter Four . Specifically, they asked for more detail regarding quantification metrics used to measure the segmentation performance of deep neural networks (DNN). Recall that the Dice similarity coefficient ( a.k.a Dice score) was used to to quantify how closely FCN-8s matched the training dataset’s hand annotated ground truth segmentation. The FCN-8s model using the Adam adaptive optimizer had a Dice score of over 96% when trained on preprocessed CT images. But

A die simulator generates a random number from 1 to 6 for each roll. You introduced a constraint to the generator such that it cannot roll the number i more than rollMax[i] (1-indexed) consecutive times. Given an array of integers rollMax and an integer n , return the number of distinct sequences that can be obtained with exact n rolls. Two sequences are considered different if at least one element differs from each other. Since the answer may be too large, return it modulo 10^9 + 7 . Example 1: Input: n = 2, rollMax = [1,1,2,2,2,3] Output: 34 Explanation: There will be 2 rolls of die, if

V1.0 模拟抛掷1个骰子，并输出其结果 案例描述： 通过计算机程序模拟抛掷骰子，并显示各点数的出现次数及频率 比如，抛掷2个骰子50次，出现点数为7的次数是8，频率是0.16 案例分析： 如何通过Python模拟随机事件？或者生成随机数？ random模块 遍历列表时，如何同时获取每个元素的索引号及其元素值？ enumerate()函数 上机实验： 1 #!/usr/bin/env python 2 # -*- coding:utf-8 -*- 3 # author: Kevin.Wang 4 # time : 2018/10/9 5 6 7 """ 8 作者：王鑫正 9 版本：1.0 10 日期：2018/10/9 11 功能： 12 """ 13 import random 14 15 16 def roll_dice(): 17 """ 18 模拟掷骰子 19 """ 20 roll = random.randint(1, 6) 21 return roll 22 23 24 def main(): 25 """ 26 主函数 27 """ 28 total_time = 10 29 30 # 初始化列表[0,0,0,0,0,0] 31 result_list = [0] * 6 32 33 for i in range(total_time): 34 roll =

dp[i][j]表示前i个骰子到达数字总和j的方案数 dp[i][j] = Σdp[i-1][j-k]，其中k是一个骰子能掷出的范围 1 #define _for(i,a,b) for(int i = (a);i < b;i ++) 2 3 class Solution 4 { 5 public: 6 int dp[31][1003]; 7 int numRollsToTarget(int d, int f, int target) 8 { 9 _for(k,1,f+1) 10 dp[0][k] = 1; 11 _for(i,1,d) 12 _for(j,1,target+1) 13 _for(k,1,f+1) 14 { 15 if(j-k>0) 16 dp[i][j] += dp[i-1][j-k]; 17 dp[i][j] %= 1000000007; 18 } 19 return dp[d-1][target]; 20 } 21 }; 来源： https://www.cnblogs.com/Asurudo/p/11334549.html