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  • R语言使用coin包应用于独立性问题的置换检验(permutation tests)、使用普通cor.test函数和置换近似spearman...
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    2022-01-17 10:31:16

    R语言使用coin包应用于独立性问题的置换检验(permutation tests)、使用普通cor.test函数和置换近似spearman_test函数、检验变量的相关性的显著性(correlation significance)

    目录

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  • Coin Test时间限制:3000ms | 内存限制:65535KB难度:1描述As is known to all,if you throw a coin up and let it droped on the desk there are usually three results. Yes,just believe what I say ~it can be ...

    Coin Test

    时间限制:3000 ms  |  内存限制:65535 KB

    难度:1

    描述

    As is known to all,if you throw a coin up and let it droped on the desk there are usually three results. Yes,just believe what I say ~it can be the right side or the other side or standing on the desk, If you don't believe this,just try In the past there were some famous mathematicians working on this .They repeat the throwing job once again. But jacmy is a lazy boy.He is busy with dating or playing games.He have no time to throw a single coin for 100000 times. Here comes his idea,He just go bank and exchange thousands of dollars into coins and then throw then on the desk only once. The only job left for him is to count the number of coins with three conditions.

    He will show you the coins on the desk to you one by one. Please tell him the possiblility of the coin on the right side as a fractional number if the possiblity between the result and 0.5 is no larger than 0.003. BE CAREFUL that even 1/2,50/100,33/66 are equal only 1/2 is accepted ! if the difference between the result and 0.5 is larger than 0.003,Please tell him "Fail".Or if you see one coin standing on the desk,just say "Bingo" any way.

    输入

    Three will be two line as input.

    The first line is a number N(1

    telling you the number of coins on the desk.

    The second line is the result with N litters.The letter are "U","D",or "S","U" means the coin is on the right side. "D" means the coin is on the other side ."S" means standing on the desk.

    输出

    If test successeded,just output the possibility of the coin on the right side.If the test failed please output "Fail",If there is one or more"S",please output "Bingo"

    样例输入

    6

    UUUDDD

    样例输出

    1/2

    英文看不懂,有道,谷歌翻译一下,看数据大致能懂,优秀代码推荐:

    [cpp]

    #include "cstdio"

    int u,d;

    int gcd(int a,int b)

    {

    if(a==0) return b;

    else return gcd(b%a,a);

    }

    int main()

    {

    int n;

    char c;

    scanf("%d",&n);

    getchar();

    for(int i=0;i!=n;i++)

    {

    c=getchar();

    if(c=='S') {puts("Bingo");return 0;}

    if(c == 'U') ++u;

    else ++d;

    }

    int g=gcd(u,u+d);

    if((double)u/(u+d)-0.5>0.003 ||(double)u/(u+d)-0.5

    展开全文
  • COIN(欧拉函数

    2019-09-29 16:01:43
    freopen("coin.out","w",stdout); 19 cin>> x; 20 ll up= sqrt(x); 21 e(up+ 1 ); 22 for (ll i= 1 ; i; i++ ){ 23 ans+=euler[i]*floor(x/(i* i)); 24 } 25 cout ans; 26 // fclose(stdin); ...

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cmath>
     4 using namespace std;
     5 typedef long long ll;
     6 ll x, ans=0;
     7 int euler[10000005];
     8 void e(int n){
     9     euler[1]=1;
    10     for(int i=2; i<=n; i++) euler[i]=i;
    11     for(int i=2; i<=n; i++)
    12         if(euler[i]==i)
    13                for(int j=i; j<n; j+=i)
    14                   euler[j]=euler[j]/i*(i-1);
    15 }
    16 int main(){
    17     //freopen("coin.in","r",stdin);
    18     //freopen("coin.out","w",stdout);
    19     cin>>x;
    20     ll up=sqrt(x);
    21     e(up+1);
    22     for(ll i=1; i<=up; i++){
    23         ans+=euler[i]*floor(x/(i*i));
    24     }
    25     cout<<ans; 
    26     //fclose(stdin);
    27     //fclose(stdout);
    28     return 0;
    29 }

     

    转载于:https://www.cnblogs.com/Aze-qwq/p/9409536.html

    展开全文
  • 比特币的实用程序(读取器/写入器、哈希函数、base58 等) 基于 许可麻省理工学院
  • @functools.lru_cache——进行函数执行结果备忘,显著提升递归函数执行时间。 示例:寻找宝藏。在一个嵌套元组tuple或列表list中寻找元素’Gold Coin’ import time from functools import lru_cache def find_...
  • 原始函数也可以使用非线性成本,该成本应适用于分段线性凸函数。 CLP还包括用于解决LP的障碍方法。 Clp是用C ++编写的,并在。NET下作为开源发布。 它是在 Clp开发站点为。 引用 当前构建状态 下载 Docker镜像 有一...
  • HDU-2069 Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent....For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cen

    HDU-2069

    Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.

    For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.

    Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.

    n ≤ 250 n ≤ 250 n250

    input

    11
    26
    76
    145
    239

    output

    4
    13
    134
    828
    3140

    code

    //Siberian Squirrel
    //#include<bits/stdc++.h>
    #include<unordered_map>
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    
    #define ACM_LOCAL
    
    using namespace std;
    typedef long long ll;
    
    const double PI = acos(-1);
    const double eps = 1e-7;
    const int inf = 0x3f3f3f3f;
    const int MOD = 3221225473;
    const int N = 8e3 + 10;
    const int UP = 101;
    
    int p[6] = {1, 50, 25, 10, 5, 1};
    int f1[N][UP], f2[N][UP];
    
    inline ll solve(int n, ll res = 0) {
        memset(f1, 0, sizeof(f1));
        f1[0][0] = 1;
        for(int i = 1; i <= 5; ++ i) {
            for(int j = 0; j <= n; ++ j) {
                for(int k = 0; j + k * p[i] <= n && k <= 100; ++ k) {
                    for(int l = 0; l <= 100 && l + k <= 100; ++ l) {
                        f2[j + k * p[i]][l + k] += f1[j][l];
                    }
                }
            }
            for(int j = 0; j <= n; ++ j) {
                for(int k = 0; k <= 100; ++ k) {
                    f1[j][k] = f2[j][k];
                    f2[j][k] = 0;
                }
            }
        }
        for(int i = 1; i <= 100; ++ i) res += f1[n][i];
        return n? res: 1;
    }
    
    int main() {
    #ifdef ACM_LOCAL
    //    freopen("input", "r", stdin);
        freopen("output", "w", stdout);
    #endif
        int o = 1, n, m, v;
    //    scanf("%d", &o);
        while(o --) {
            while(~scanf("%d", &n)) {
                printf("%lld\n", solve(n));
            }
        }
        return 0;
    }
    
    展开全文
  • 总母函数的角度:同样相当于取5种类型的硬币,然后通过母函数性质模拟多项式乘积 便可以得到 多重背包完整代码: #include #include #include using namespace std; int a[ 255 ][ 105 ],b[ 255 ]...
  • 题意:给你50 25 10 5 1硬币各一些 问你组成某个数的不同方案数 但是注意 硬币的个数不超过一百个 所以这题要控制硬币的个数 用二维数组控制 AC代码: #include ...int f[6] ={0,1,5
  • Coin和Token的区别

    千次阅读 2019-01-06 10:31:27
    所有的coin和token都被认为是密码货币,尽管事实上它们中有许多不是以货币形式流通,而且从来就不是那样的。根据定义,货币是交换媒介、记账单位或价值储存。比特币具有这些特性,因此在这种情况下,“加密货币”这...
  • 上次学会了用母函数来解决这种排列组合的问题,发现这一题也可用母函数来解决,不过要解决如何限制硬币数不超过100。经过一番测试,发现可以用二维数组来做,行的值表示硬币的和值,列的值表示和值为行的值的硬币的...
  • Coin3D三维可视化教程2

    2020-05-03 21:18:38
    这个是在Coin3D三维可视化教程1基础上的继续。 上次介绍了SoRotationXYZ 与时间SoElapsedTime绑定实现动态旋转,这次实现手动旋转的方式。方法是在圆锥的外面添加一个操作器(轨迹球)SoTrackballManip。轨迹球本身...
  • COIN(语境推断)模型[ ]是运动学习的原则贝叶斯模型,其中针对不同的上下文获取了单独的记忆。 该模型的关键见解是,内存的创建,更新和表达都由单个计算(上下文推断)控制。 该模型是在MATLAB中开发的,并已在...
  • R语言使用coin包应用于独立性问题的置换检验(permutation tests)、对于成对的数据,使用wilcox.test函数进行两组数据的Wilcoxon符号秩检验、(Wilcoxon signed rank)、使用wilcoxsign_test函数执行Wilcoxon符号秩...
  • Coin ChangeTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22298 Accepted Submission(s): 7801Problem DescriptionSuppose there are 5 types of ...
  • Coin Change Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16135 Accepted Submission(s): 5485 Problem Description Suppose there
  • Coin Change Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.  For example, if we have 11 cents...
  • Coin Change Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 20457 Accepted Submission(s): 7201 Problem Description Suppose there
  • PHP调用未定义的函数

    2021-03-22 22:09:58
    如何重现错误,以及如何解决它:>将此代码放在一个名为p.php的文件中:class yoyo{function salt(){}function pepper(){salt();}}$y = new yoyo();$y->pepper();?>>运行如下:php p.php>...
  • R语言使用coin包应用于分类变量独立性问题的置换检验(permutation tests)、使用普通卡方检验chisq.test函数和置换近似卡方检验chisq.test函数、检验分类变量的独立性(Chi-square test of independence)
  • CoinChange问题CoinChange问题是很经典的问题,在leetcode上有518. Coin Change 2 h和322. Coin Change 。其中: Coin Change 2:给定固定有限种类a1,a2,...,ama_1,a_2,...,a_m的硬币,求换成金额nn的方法数,即求...
  • 零钱兑换(coin-change) 动态规划问题

    千次阅读 2019-12-01 20:43:14
    文章目录零钱兑换(coin-change)...编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额,返回 -1。 示例 1: 输入: coins = [1, 2, 5], amount = 11 输出: 3 解释: 11 = 5...
  • #include<iostream>#include<cstdio>#include<cmath>#include<vector>#include<cstdlib>using namespace std; int main(){int dp[260][105];int a[5]={50,25,10,5,1};......
  • 思路:组合问题,可以考虑用母函数,但是这里考虑组合的个数和组合的最大数。硬币的个数不超过100;而且硬币的 总值不大于250。要同时兼顾这些问题,同一总值,可以由不同个数的硬币组成,即相互对应的关系,就得...
  • Coin3D三维可视化教程1

    千次阅读 2020-05-03 21:05:21
    上次介绍了Coin3D的安装和在VS 2019 +QT下的配置,后面讲逐步学习这个库的使用,采用的主要材料的The Inventor Mentor中文版。书中的代码时在Mac平台的开发,这里采用的用的是Windows,SoXt是Coin3D在Mac平台上的...
  • dp[i][j]表示前i种硬币组成j元的方案数。 状态转移方程:dp[i][j] = ...也可以用母函数写,注意母函数只能预处理的时候跑完,不然超时。 递推: #include #include #include using namespace std; typedef long lon
  • Coin Change Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9842Accepted Submission(s): 3300 Problem Description Suppose there are...
  • Coin3D三维可视化教程7

    2020-05-05 11:31:49
    当构造SoGLRenderAction (见第 9 章)对象时,需要指定视口作为构造函数的其中的一个参数。 SoCamera的viewportMapping域是用来指定当照相机的横纵比和视口不同时,如何将照相机的投影映射到视口上。前三个选项是通过...
  • 第二章-Coin Dash

    2020-11-21 20:17:58
    Coin DashProject setup向量和2D坐标系VectorsPixel rendering像素渲染第一部分–玩家场景创建场景精灵动画碰撞形状编写Player脚本移动Player关于delta选择动画开始和结束玩家的移动准备碰撞第2部分-硬币场景节点...

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