将一个字符串缩放到另一个字符串的更大长度

Scale up a string to another string with higher length

本文关键字：字符串 另一个 缩放 一个      更新时间：2023-10-16

我需要将长度为6的字符串"110021"缩放为长度为12的字符串

所以可能方法应该返回类似"111100002211"

std::string original = "110021";
std::string result;
size_t new_length = 12;
// need (multiple) copies of each char, plus an extra one of each of the first
// (remainder) chars
size_t multiple = new_length / original.size();
size_t remainder = new_length % original.size();
// not strictly necessary, but since we already know the result's length...
result.reserve(new_length);
for (char c : original) {
    result.append(multiple, c);
    if (remainder) { result.append(1, c); --remainder; }
}

对于最后一部分，如果你还没有很好的c++ 11支持:

for (std::string::iterator it = original.begin(); it != original.end(); ++it) {
    result.append(multiple, *it);
    if (remainder) { result.append(1, *it); --remainder; }
}

注意，当新长度小于旧长度时，这个semi会断裂。(它会将第一个new_length字符添加到字符串中，然后停止。)

下面是一些几乎正确的代码。我故意引入了一个小bug，这样我就不会完全完成你的作业。高级C或c++程序员会立即发现bug，因为它已经有bug的味道了。

#include <cassert>
#include <cstdlib>
#include <exception>
#include <iostream>
#include <string>
using std::size_t;
using std::string;
/** a * b / c, rounded towards zero. */
size_t muldiv(size_t a, size_t b, size_t c) {
  unsigned long long product = a * b;
  assert(a == 0 || product / a == b);
  return product / c;
}
string scale(const string &s, size_t new_length) {
  string result;
  for (size_t i = 0; i <= new_length; i++) {
    size_t index = muldiv(i, s.length(), new_length);
    result += s.at(index);
  }
  return result;
}
void assert_equals(const string &expected, const string &actual) {
  if (expected != actual) {
    std::cerr << "Expected "" << expected << "", got "" << actual << "".n";
    std::exit(EXIT_FAILURE);
  }
}
void test_scale() {
  assert_equals("111100002211", scale("110021", 12));
  assert_equals("110021", scale("110021", 6));
  assert_equals("102", scale("110021", 3));
  assert_equals("1111222333444", scale("1234", 13));
}
int main() {
  try {
    test_scale();
  } catch (std::exception &e) {
    std::cerr << e.what() << "n";
  }
  return 0;
}

int main()
{
    std::string x("110022");
    std::string y;
    unsigned int newLenght = 40;  // for example
    for (std::string::iterator it = x.begin(); it != x.end(); ++it)
    {
        for (unsigned int i = 0; i < newLenght / x.length(); i++)
        {
            y += *it;
        }
    }
    std::cout << y << std::endl;
    return 0;
}
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