c – 获取长度字符串的范围百分比

发布时间：2020-12-16 09:58:41 所属栏目：百科来源：网络整理

导读：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一定的手段范围;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,A7

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一定的手段范围;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";

弦手的长度是662;为了获得范围,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％.

这很难,因为弦不能在任何地方切割,好的输出应该是例如72o,有或没有逗号到底,这不是最重要的.

我试着像这样创建一个子串：

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％,而且切割很差,输出为：AA,A
应该是：AA,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……

问候,
GUI

解决方法

这是一种方法：获取字符串,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.

这就是我在Python中如何做到这一点：

start,stop = 30,60
hands = 'AA,
ATo,
KJo,
K3s,
T9o,
T6s,
54o,
Q3o,
J5o,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,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;
}

（编辑：李大同）

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