获取长度字符串的范围百分比

Question

我遇到了问题。我正在学习 C++。我采用了扑克环境； 目标是从字符串长度中得到一定的手牌范围；该字符串代表扑克游戏中的所有起手牌，顺序为强手。

 string hand = "AA,KK,QQ,JJ,AKs,AKo,AQs,AQo,TT,AJs,ATs,AJo,KQs,KJs,KTs,QJs,ATo,QTs,JTs,A9s,A9o,KQo,A8s,A8o,A7s,A7o,A6s,A6o,A5s,A5o,A4s,99,A4o,A3s,A2s,KJo,J9s,T9s,Q9s,QJo,KTo,Q9o,88,77,66,QTo,A3o,A2o,JTo,K9s,K8s,K7s,K6s,K5s,K4s,K3s,K2s,Q8s,Q7s,Q6s,Q5s,K9o,J8s,T8s,98s,97s,87s,86s,76s,96s,75s,65s,64s,J9o,T9o,55,54s,53s,52s,K8o,43s,32s,42s,J7s,T7s,K7o,44,33,22,Q4s,Q3s,Q2s,J6s,J5s,T6s,T5s,J4s,K6o,Q8o,J8o,T8o,98o,97o,87o,85s,K5o,K4o,K3o,K2o,95s,74s,76o,65o,54o,84s,94s,Q7o,J7o,T7o,Q6o,J3s,T4s,J2s,Q5o,T3s,T2s,Q4o,J6o,86o,T6o,96o,93s,Q3o,74o,84o,75o,64o,T2o,94o,53o,93o,63o,43o,92o,73o,83o,52o,82o,42o,62o,72o,J5o,63s,92s,73s,Q2o,J4o,83s,85o,82s,T5o,95o,J3o,62s,T4o,J2o,72s,T3o,32o";

弦手的长度为662；为了获得范围，我们需要排除一定比例的起始手牌并排除一定比例的结束手牌。

因此，如果我需要最后 10% 的牌，我将排除前 90% 的牌。如果我需要 30% 到 60% 之间，我将排除前 29% 和后 39% 。

这很难，因为字符串不能在任何地方剪切，一个好的输出应该是例如 72o,J5o,63s,92s,73s,Q2o,J4o,83s, 有或没有逗号到最后，这不是最重要的。

我试图创建一个这样的子字符串：

  int startpos = 0;
  int stoppos= (662 * 7) / 100;
  string str2 = hand.substr(startpos, stoppos);
  cout <<  str2 << endl;

但这不是范围问题的答案。它只得到第一个 X%，而且切割不好，输出是：AA,KK,QQ,JJ,A 应该是：AA,KK,QQ,JJ,AKs

我为此花了很多时间。我乐于寻求建议、指导，甚至是解决方案 ..

此致，贵

Answer 1

这是一种方法：获取您的字符串，将其转换为序列，丢弃您不感兴趣的序列部分，然后将其转换回字符串。

下面是我在 Python 中的做法：

start, stop = 30, 60
hands = 'AA,KK,QQ,JJ,AKs,AKo,AQs,AQo,TT,AJs,ATs,AJo,KQs,KJs,KTs,QJs,\
ATo,QTs,JTs,A9s,A9o,KQo,A8s,A8o,A7s,A7o,A6s,A6o,A5s,A5o,A4s,99,A4o,A3s,A2s,\
KJo,J9s,T9s,Q9s,QJo,KTo,Q9o,88,77,66,QTo,A3o,A2o,JTo,K9s,K8s,K7s,K6s,K5s,K4s,\
K3s,K2s,Q8s,Q7s,Q6s,Q5s,K9o,J8s,T8s,98s,97s,87s,86s,76s,96s,75s,65s,64s,J9o,\
T9o,55,54s,53s,52s,K8o,43s,32s,42s,J7s,T7s,K7o,44,33,22,Q4s,Q3s,Q2s,J6s,J5s,\
T6s,T5s,J4s,K6o,Q8o,J8o,T8o,98o,97o,87o,85s,K5o,K4o,K3o,K2o,95s,74s,76o,65o,\
54o,84s,94s,Q7o,J7o,T7o,Q6o,J3s,T4s,J2s,Q5o,T3s,T2s,Q4o,J6o,86o,T6o,96o,93s,\
Q3o,74o,84o,75o,64o,T2o,94o,53o,93o,63o,43o,92o,73o,83o,52o,82o,42o,62o,72o,\
J5o,63s,92s,73s,Q2o,J4o,83s,85o,82s,T5o,95o,J3o,62s,T4o,J2o,72s,T3o,32o'

hands = hands.split(',')
hands = hands[start * len(hands) / 100 : stop * len(hands) / 100]
print ','.join(hands)

在两个 Whosebug 线程 here and here 的帮助下，我将 Python 代码翻译成（诚然不是惯用的）C++：

#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>

using namespace std;

vector<string> &split(const string &s, char delim, vector<string> &elems) {
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) {
    elems.push_back(item);
  }
  return elems;
}

vector<string> split(const string &s, char delim) {
  vector<string> elems;
  split(s, delim, elems);
  return elems;
}

int main() {
  // percentage markers to keep, 30% -- 60% here
  int start = 30, stop = 60;
  string hands = "AA,KK,QQ,JJ,AKs,AKo,AQs,AQo,TT,AJs,ATs,AJo,KQs,KJs,KTs,QJs,\
ATo,QTs,JTs,A9s,A9o,KQo,A8s,A8o,A7s,A7o,A6s,A6o,A5s,A5o,A4s,99,A4o,A3s,A2s,\
KJo,J9s,T9s,Q9s,QJo,KTo,Q9o,88,77,66,QTo,A3o,A2o,JTo,K9s,K8s,K7s,K6s,K5s,K4s,\
K3s,K2s,Q8s,Q7s,Q6s,Q5s,K9o,J8s,T8s,98s,97s,87s,86s,76s,96s,75s,65s,64s,J9o,\
T9o,55,54s,53s,52s,K8o,43s,32s,42s,J7s,T7s,K7o,44,33,22,Q4s,Q3s,Q2s,J6s,J5s,\
T6s,T5s,J4s,K6o,Q8o,J8o,T8o,98o,97o,87o,85s,K5o,K4o,K3o,K2o,95s,74s,76o,65o,\
54o,84s,94s,Q7o,J7o,T7o,Q6o,J3s,T4s,J2s,Q5o,T3s,T2s,Q4o,J6o,86o,T6o,96o,93s,\
Q3o,74o,84o,75o,64o,T2o,94o,53o,93o,63o,43o,92o,73o,83o,52o,82o,42o,62o,72o,\
J5o,63s,92s,73s,Q2o,J4o,83s,85o,82s,T5o,95o,J3o,62s,T4o,J2o,72s,T3o,32o";

  vector<string> hands_ = split(hands, ',');
  start = start * hands_.size() / 100;
  stop = stop * hands_.size() / 100;

  stringstream buffer;
  for (size_t i = start; i < stop; ++i) {
    if(i != start)
      buffer << ',';
    buffer << hands_[i];
  }
  string output = buffer.str();

  cout << output << endl;
  return 0;
}

Answer 2

另一种方法：找到我们感兴趣的子字符串的索引，并提取该子字符串。不过也没有短多少。

#include <algorithm>
#include <iostream>
#include <string>

using namespace std;

int main() {
  // percentage markers to keep, 30% -- 60% here
  int start = 30, stop = 60;
  string hands = "AA,KK,QQ,JJ,AKs,AKo,AQs,AQo,TT,AJs,ATs,AJo,KQs,KJs,KTs,QJs,\
ATo,QTs,JTs,A9s,A9o,KQo,A8s,A8o,A7s,A7o,A6s,A6o,A5s,A5o,A4s,99,A4o,A3s,A2s,\
KJo,J9s,T9s,Q9s,QJo,KTo,Q9o,88,77,66,QTo,A3o,A2o,JTo,K9s,K8s,K7s,K6s,K5s,K4s,\
K3s,K2s,Q8s,Q7s,Q6s,Q5s,K9o,J8s,T8s,98s,97s,87s,86s,76s,96s,75s,65s,64s,J9o,\
T9o,55,54s,53s,52s,K8o,43s,32s,42s,J7s,T7s,K7o,44,33,22,Q4s,Q3s,Q2s,J6s,J5s,\
T6s,T5s,J4s,K6o,Q8o,J8o,T8o,98o,97o,87o,85s,K5o,K4o,K3o,K2o,95s,74s,76o,65o,\
54o,84s,94s,Q7o,J7o,T7o,Q6o,J3s,T4s,J2s,Q5o,T3s,T2s,Q4o,J6o,86o,T6o,96o,93s,\
Q3o,74o,84o,75o,64o,T2o,94o,53o,93o,63o,43o,92o,73o,83o,52o,82o,42o,62o,72o,\
J5o,63s,92s,73s,Q2o,J4o,83s,85o,82s,T5o,95o,J3o,62s,T4o,J2o,72s,T3o,32o";

  int nhands = count(hands.begin(), hands.end(), ',');
  start = start * nhands / 100;     // percentage -> comma count
  stop = stop * nhands / 100;       // percentage -> comma count

  int start_ = start, stop_ = stop; // copies
  int i = 0, substr_start, substr_end;

  for (i = 0; i < hands.length() && start_ > 0; ++i) {
    if (hands[i] == ',')
      start_--;
  }
  substr_start = i;

  stop_ -= start - 1;               // already counted 'start' commas
  for (; i < hands.length() && stop_ > 0; ++i) {
    if (hands[i] == ',')
      stop_--;
  }
  substr_end = i;

  cout << hands.substr(substr_start, substr_end - substr_start - 1) << endl;
  return 0;
}

Answer 3

我只是想指出，弦的百分比与手的百分比不同。

不同的手或多或少比其他人更常见。具体来说，非同花牌（例如 AKo）出现的可能性是同花版本 (AK) 的三倍，而对子的可能性是非同花牌的一半。这是因为从一副牌中抽出牌的组合学。

对于配对，每个等级有 C(4,2) = 6 种可能的组合。

对于同花牌，只有花色不同，所以有 4 种可能的组合。

对于非同花牌，第一张牌有四种可能的花色，但第二张牌只有三种花色（否则就是同花），所以有 12 种组合。

为了获得正确的百分比，根据发牌的可能性，您需要相应地加权。

获取长度字符串的范围百分比

Get range percentage of a length string

c++

poker

substring