获取长度字符串的范围百分比

时间：2015-07-18 18:45:45

标签： c++ substring poker

我遇到了问题。我正在学习c ++。我拿了一个扑克背景; 目标是从字符串长度获得某个手牌范围;该字符串表示扑克游戏中的所有起手牌，顺序较强。

 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";

字符串手的长度为662 ;要获得范围，我们需要排除起手牌的百分比并排除结束牌的百分比。

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

这很困难，因为字符串不能在任何地方剪切，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。

我尝试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％，而且剪切很糟糕，输出为：AA,KK,QQ,JJ,A 应该是：AA,KK,QQ,JJ,AKs

我花了很多时间在这上面。我愿意接受建议，指导，甚至解决方案..

此致 GUI

3 个答案:

答案 0 :(得分：4)

这是一种方法：取你的字符串，将其转换为序列，丢弃你不感兴趣的序列部分，然后将其转换回字符串。

所以我在Python中如何做到这一点：

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Stackoverflow线程here和here的帮助下，我将Python代码翻译成（当然不是惯用语）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;
}

答案 1 :(得分：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。但不会短得多。

#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;
}

答案 2 :(得分：1)

我只想指出字符串的百分比与指针的百分比不同。

不同的手或多或少比其他人更常见。具体而言，Offsuit指针（例如AKo）比适合的版本（AK）发生的可能性高三倍，并且成对的可能性是手上的一半。这是因为从卡片中拉出卡片的组合。

对于对，每个等级有C（4,2）= 6种可能的组合。

对于合适的手，只有套装会有所不同，因此有4种可能的组合。

对于非法手套，第一张卡可以有四种套装，但第二种卡只有三套（否则它适合），所以有12种组合。

为了获得适当的百分比，就手的可能性而言，你需要相应地对手进行加权。

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