uva1593—— Alignment of Code 【stl,模拟】

本文介绍了一种将代码中的定义或其他表格信息进行垂直对齐的方法,确保每列单词都整齐排列,同时保持代码尽可能紧凑。文章详细解释了如何通过读取和格式化输入文件来实现这一目标,并提供了一个具体的代码实现示例。

You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before position p2 − 2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3 − 2, etc. For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).

Input

The input file contains one or more lines of the code up to the end of file. All lines (including the last one) are terminated by a standard end-of-line sequence in the file. Each line contains at least one word, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the input file.

Output

Write to the output file the reformatted, aligned code that consists of the same number of lines, with the same words in the same order, without trailing and leading spaces, separated by one or more spaces such that i-th word on each line starts at the same position pi . Note for the Sample: The ‘⊔’ character in the example below denotes a space character in the actual files (ASCII code 32).

Sample Input 

␣␣start:␣␣integer;␣␣␣␣//␣begins␣here

stop:␣integer;␣//␣␣ends␣here

␣s:␣␣string;

c:␣␣␣char;␣//␣temp 

Sample Output

start:␣integer;␣//␣begins␣here

stop:␣␣integer;␣//␣ends␣␣␣here

s:␣␣␣␣␣string;

c:␣␣␣␣␣char;␣␣␣␣//␣temp

题目大意:给你一些字符,让你把这组字符按样例所给形式对其输出。

大致思路:我们可以用vector储存一个单词矩阵,然后找单词矩阵每一列的最大长度,格式化输出就行了。

首先就是这道题的输入问题,我们可以getline把字符按行输入,在用stringstream最字符进行格式化存入vector中。最后输出的时候我们要调用#include <iomanip> 中的setw函数进行格式化输出。

至于setw,和stringstream的应用可以参考一下两片博客。

stringstream的使用

setw的使用 

接下来就是代码了:

#include <iomanip>
#include <bits/stdc++.h>
using namespace std;
vector<string> word[100010];
int maxlen[100010];
int main(){
    string s,t;
    int k = 0;
    memset(maxlen,0,sizeof(maxlen));
    while(getline(cin,s)){
        //按行读入每一个字符
        stringstream ss(s); //把每一个字符转化为标准形式
        //string t;
       // int k = 0;
        int i = 0;
        while(ss >> t){
           // cout<<t<<endl;
            maxlen[i] = max((int)t.length(),maxlen[i]);
            word[k].push_back(t);
           // cout<<t<<endl;
            //printf("%d\n",k);
            i++;
        }
        k++;
    }
  //  printf("%d\n",k);
    cout << setiosflags(ios::left);
    for(int i = 0; i < k; i++){
        int j;
        for(j = 0; j < (int)word[i].size()-1; j++){
            cout<<setw(maxlen[j] + 1)<<word[i][j];
        }
        cout << word[i][j] <<endl;
    }
    return 0;
}

 

在编程或系统设计中,出现错误信息“increases required alignment of target type”通常与内存对齐(memory alignment)有关。这类问常见于C/C++等底层语言,尤其是在处理结构体、联合体或特定硬件平台(如嵌入式系统)时[^1]。 ### 错误含义 该错误提示表明某个操作试图将一个类型转换为另一个具有更严格对齐要求的目标类型。例如,如果一个指针原本指向的是按4字节对齐的数据类型(如int),而你试图将其转换为按8字节对齐的类型(如double),就可能触发此错误。这种情况下,编译器会认为当前的指针或数据布局无法满足目标类型的对齐需求,从而拒绝编译或运行时操作[^2]。 ### 常见场景 - **强制类型转换**:将void*或其他通用指针转换为具体类型时。 - **结构体内存布局**:结构体成员之间可能存在填充(padding),导致整体大小不一致。 - **跨平台开发**:不同架构(如x86 vs ARM)对数据类型的对齐要求不同。 ### 解决方案 #### 1. 显式对齐内存分配 使用`aligned_alloc`或`std::aligned_alloc`来确保分配的内存满足特定类型的对齐要求: ```cpp #include <cstdlib> void* ptr = aligned_alloc(alignof(double), sizeof(double)); ``` #### 2. 使用`std::aligned_storage` 当需要手动管理内存但希望支持多种类型时,可以使用`std::aligned_storage`来创建足够大且对齐的缓冲区: ```cpp #include <type_traits> union { std::aligned_storage<sizeof(double), alignof(double)>::type storage; double d; } u; new (&u.d) double(3.14); // 安全构造 ``` #### 3. 结构体对齐控制 在结构体定义中,可以通过`#pragma pack`或`[[gnu::packed]]`等指令控制成员的对齐方式,但这可能导致性能下降或访问异常: ```cpp #pragma pack(push, 1) struct PackedStruct { char c; int i; }; #pragma pack(pop) ``` #### 4. 避免不当的指针转换 确保指针转换前后的类型对齐要求一致,或者使用`reinterpret_cast`时明确了解其后果: ```cpp int* pInt = new int(42); double* pDouble = reinterpret_cast<double*>(pInt); // 潜在对齐错误 ``` 应改为: ```cpp double value = static_cast<double>(*pInt); ``` #### 5. 使用`std::launder`(C++17及以上) 当涉及对象生命周期和别名规则时,使用`std::launder`可帮助编译器理解新对象的布局: ```cpp #include <memory> auto raw = malloc(sizeof(double)); double* p = new (raw) double(3.14); double* pCorrect = std::launder(p); ``` --- ### 总结 “increases required alignment of target type”错误本质上是由于类型转换过程中未能满足目标类型的内存对齐要求。解决方法包括显式对齐内存分配、使用标准库提供的对齐工具、避免不当的指针转换以及合理控制结构体成员的对齐方式。通过这些手段,可以在保证程序正确性的同时提升性能和可移植性。 ---
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