C ++：如何分割字符串成均匀的尺寸较小的字符串？ [英] C++: How do I split a string into evenly-sized smaller strings?

问题描述

在C ++中，我怎么字符串分割成均匀的尺寸更小的字符串？

In C++, how do I split a string into evenly-sized smaller string?

例如，我有一个字符串012345678，并希望将其分成5小串，这应该返回我像01，23，45，67，8的东西。

For example, I have a string "012345678" and want it to split it into 5 smaller strings, and this should return me something like "01", "23", "45", "67", "8".

我有确定的较小的字符串长度的麻烦。在previous例如，原来的字符串大小9，我想把它分成5个较小的字符串，因此除了最后一个每个小字符串应该是长度为9/5 = 1，但最后一个会长度9 - 1 * 4 = 5，这是不能接受的

I'm having trouble of determining the length of the smaller strings. In the previous example, the original string is size 9, and I want to split it into 5 smaller string, so each smaller string except the last one should be length 9 / 5 = 1, but then the last one will be length 9 - 1* 4 = 5, which is unacceptable.

所以这个问题的正式定义：原始的字符串被分为n个子，并没有两个子应在长度大于1不同。

So the formal definition of this problem: the original string is split into EXACTLY n substrings, and no two of the substrings should differ by greater than 1 in length.

我的重点不是C ++语法或图书馆。它是如何设计一个算法，使返回的字符串可以是几乎相等的大小。

My focus is not on C++ syntax or library. It's how to design an algorithm so that the returned string can be nearly-equal in size.

推荐答案

要<一href="http://stackoverflow.com/questions/8084010/algorithm-for-subdividing-an-array-into-semi-equal-uniform-sub-arrays/8084403#8084403">divide N个项目成M部分的，具有长度在一个单元内，可以使用公式（N * I + N）/ M - （N * I）/ M 作为我长度个部分，如下图所示。

To divide N items into M parts, with lengths within one unit, you can use formula (N*i+N)/M - (N*i)/M as length of i'th part, as illustrated below.

 #include <string>
 #include <iostream>
 using namespace std;

 int main() {
   string text = "abcdefghijklmnopqrstuvwxyz";
   int N = text.length();
   for (int M=3; M<14; ++M) {
     cout <<" length:"<< N <<"  parts:"<< M << "\n";
     int at, pre=0, i;
     for (pre = i = 0; i < M; ++i) {
       at = (N+N*i)/M;
       cout << "part " << i << "\t" << pre << "\t" << at;
       cout << "\t" << text.substr(pre, at-pre) << "\n";
       pre = at;
     }
   }
   return 0;
 } 

例如，当 M 是4或5，code以上生产：

For example, when M is 4 or 5, the code above produces:

  length:26  parts:4
 part 0 0   6   abcdef
 part 1 6   13  ghijklm
 part 2 13  19  nopqrs
 part 3 19  26  tuvwxyz
  length:26  parts:5
 part 0 0   5   abcde
 part 1 5   10  fghij
 part 2 10  15  klmno
 part 3 15  20  pqrst
 part 4 20  26  uvwxyz

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