• 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）

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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
• 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

11
26
76
145
239

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
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• 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...
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