Contestant and Gladiator problem

jaka+realsense+aruco+手眼标定(python) 第二次重建了项目的手眼标定部分……有很多第一次记录遗漏的地方,在此补充。 阅读详情

You are consulting for a game show in which n contestants are pitted against n gladiators in order to see which contestants are the best. The game show aims to rank the contestants in order of strength; this is done via a series of 1-on-1 matches between contestants and gladiators. If the contestant is stronger than the gladiator, then the contestant wins the match; otherwise, the gladiator wins the match. If the contestant and gladiator have equal strength, then they are “perfect equals” and a tie is declared. We assume that each contestant is the perfect equal of exactly one gladiator, and each gladiator is the perfect equal of exactly one contestant. However, as the gladiators sometimes change from one show to another, we do not know the ordering of strength among the gladiators.

The game show currently uses a round-robin format in which matches are played and contestants are ranked according to their number of victories. Since few contestants can happily endure gladiator confrontations, you are called in to optimize the procedure.

You are asked to give a reasonable representation for the input of the problem and give a randomized algorithm for ranking the contestants, whose expected number of matches should be . Besides the original program, a detailed design reported is required.

 

[code]

#include <iostream.h>
#include <stdlib.h>
#include <stdio.h>
#include <time.h>

int Rand(int p,int r){
    time_t t;
    srand((unsigned) time(&t));
    return  p + (rand() % (r-p+1));
}

void swap(int &i,int &j)
{
 int tmp = i;
 i = j;
 j = tmp;
}

int Partition(int * A,int p,int q ,int b)
{
 int i = p-1;
 int k = 0;
 for(int j = p; j <= q; j++)
 {
  if(A[j] < b)
  {
   i = i+1;
   swap(A[i],A[j]);
  }
  if(A[j] == b)
   k = j;
 }
    i++;
 swap(A[i],A[k]);
 return i;
}

void QuickSort(int * C,int p,int q,int * G)
{ 
 
 if(p < q)
 { 
  int rand = Rand(p,q);
  int m = Partition(G,p,q,C[q]);
  int n = Partition(C,p,q,G[m]);
  QuickSort(C,p,n-1,G);
  QuickSort(C,n+1,q,G);
 }
}

int main()
{
 int Contestant[8] = {3,7,4,6,1,5,8,2};
 int Gladiator[8] = {6,4,8,1,3,5,2,7};
 
    QuickSort(Contestant,0,7,Gladiator);
 
 printf("The sorted list of Contestant is: /n");
 for(int k = 0; k < 8; k++)
 {
  printf("%4d",Contestant[k]);
 }
 printf("/n");
 
 printf("The sorted list of Gladiator is: /n");
 for(int l = 0; l < 8; l++)
 {
  printf("%4d",Gladiator[l]);
 }
 printf("/nThe end! /n");
 return 1;
}

[code]

Vivado 18.3安装全攻略:从下载到配置的完整指南 本文详细介绍了Vivado 18.3的安装全流程,从下载准备到配置优化,特别适合FPGA开发者。文章包含硬件要求、版本选择、组件配置等实用技巧,帮助用户避免常见安装问题,确保稳定运行。重点解析了License配置和系统优化设置,提升开发效率。 阅读详情

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