比较10个塔。修复代码 [英] Compare 10 towers.fix the code
问题描述
我的代码未通过测试
给出了十个塔。你需要比较它们。取幂从右到左发生a ^（a ^（a ^ a））。最后，按升序打印索引。这是我的代码但不正确。
输入
10 //塔楼数量
4 2 2 2 2 2 // 4一行中的第一个数字不是塔的元素，它是//中的元素数减一。
1 2 2
1 3 2
1 2 3
3 2 2 2 2
2 2 2 2
3 3
3 3 3 3
2 4 3 3
2 2 3 4
输出
2 4 3 6 7 5 9 10 1 8
< b>我尝试过：
my code fails the tests
Ten towers are given. You need to compare them. exponentiation occurs from right to left a ^ (a ^ (a ^ a)). At the end, print their indexes in ascending order. Here is my code but it is incorrect.
input
10 // number of towers
4 2 2 2 2 2 // 4 The first number in a line is not an element of the tower, it is the //number of elements in it minus one.
1 2 2
1 3 2
1 2 3
3 2 2 2 2
2 2 2 2
1 3 3
3 3 3 3 3
2 4 3 3
2 2 3 4
output
2 4 3 6 7 5 9 10 1 8
What I have tried:
#include <fstream>
#include <algorithm>
#include <stdio.h>
#include <math.h>
using namespace std;
class tower_t {
public:
int num; // the tower room
int height; // the height of the tower
double val[11]; // content
double cache[11]; // cache to speed up the calculation
// Designer
tower_t() {
for (int i = 0; i < 11; i++) {
val[i] = 1;
cache[i] = 0;
}
height = 0;
}
// Triple logarithm of the top 3 levels
double head(int level) {
if(cache[level] == 0) cache[level] = log10(log10(val[level])) + log10(val[level + 1]) * val[level + 2];
return cache[level];
}
// The calculation of the tops until intermeddle in double
void normalize() {
while(height > 1 && (log10(val[height - 2]) * val[height - 1]) < 50) {
val[height - 2] = pow(val[height - 2], val[height - 1]);
val[height - 1] = 1;
height--;
}
}
// Output for debugging
void print() {
#ifdef _DEBUG
printf("%2d: {", num);
for (int i = 0; i < height; i++) {
if (i > 0) printf(", ");
if(val[i] < 1000000000) {
printf("%0.0f", val[i]);
} else {
printf("%0.3e", val[i]);
}
}
printf("}\n");
#endif
}
};
// comparison of two towers
bool compare(tower_t& t1, tower_t& t2) {
// floor with which to compare the last three levels
int level = ((t1.height > t2.height) ? t1.height : t2.height) - 3;
if (level < 0) level = 0;
if(t1.height == t2.height) { // if the towers are of the same height, compare by floor
for (int i = t1.height - 1; i >= 0; i--) {
if (abs(t1.val[i] - t2.val[i]) > (t1.val[i] * 1e-14)) {
if (i < level) { // the tops of the towers coincided below level
return t1.val[i] < t2.val[i];
}
break;
}
}
}
return t1.head(level) < t2.head(level);
}
int main(int argc, char**argv)
{
// Reading job
ifstream in ("input.txt");
int cnt;
in >> cnt;
tower_t* towers = new tower_t[cnt];
for (int i = 0; i < cnt; i++) {
int len;
in >> len;
towers[i].num = i + 1;
bool write = true;
for (int j = 0; j <= len; j++) {
int val;
in >> val;
if (val <= 1) write = false; // if level of <= 1 the higher not to read
if(write) {
towers[i].val[j] = val;
towers[i].height = j + 1;
}
}
towers[i].print();
towers[i].normalize();
}
// Sort
sort(towers, towers + cnt, compare);
// The output
ofstream out("output.txt");
for (int i = 0; i < cnt; i++) {
out << towers[i].num << " ";
towers[i].print();
}
out << endl;
out.close();
delete[] towers;
return 0;
}
推荐答案
编译并不意味着你的代码是对的！ ：笑：
将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。
所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。
首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：
Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.
So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.
Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input Expected output Actual output
1 2 1
2 4 4
3 6 9
4 8 16然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。
所以，你可以查看代码和很明显，它在某处：
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
int Double(int value)
{
return value * value;
}
一旦你知道可能出现的问题，就开始使用调试器找出原因。在方法的第一行放置一个断点，然后运行你的应用程序。当它到达断点时，调试器将停止，并将控制权移交给您。您现在可以逐行运行代码（称为单步执行）并根据需要查看（甚至更改）变量内容（哎呀，您甚至可以更改代码并在需要时再试一次）。 />
在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。
如果没有，为什么不呢？它有什么不同？
希望这可以帮助你找到代码的哪个部分有问题，以及问题是什么。
这是一项技能，它是一个值得开发的，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改进！
Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
你的工作也是调试你的代码。
Your job is also to debug your code.
这是我的代码，但它不正确
Here is my code but it is incorrect
您的代码不是单片，它是一组步骤，只需更改，您就可以检查中间结果。这样做，你就会知道出了什么问题。
-----
顺便说一下，你给出了一个输出，你忘了说它是否是实际输出或预期的输出，因为你没有给出其他输出我们不知道它是什么错误。
实际输出是......
预期输出是：...
-----
你的代码没有你想象的那样，你不明白为什么！
有一个几乎通用的解决方案：逐步在调试器上运行代码，检查变量。
调试器在这里显示你的代码正在做什么和你的任务是与它应该做的比较。
调试器中没有魔法，它不知道你应该做什么，它没有找到bug，它只是帮助你通过向您展示正在发生的事情。当代码没有达到预期的效果时，你就接近了一个错误。
要查看你的代码在做什么：只需设置断点并查看代码是否正常运行，调试器允许你执行第1行第1行，并在执行时检查变量。
调试器 - 维基百科，免费的百科全书 [ ^ ]
掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]
使用Visual Studio 2010进行基本调试 - YouTube [ ^ ]
1.11 - 调试程序（步进和断点）|学习C ++ [ ^ ]
调试器只显示你的代码正在做什么，你的任务是与它应该做什么进行比较。
[更新]
你在解决方案3中的例子很奇怪。
Your code is not monolithic, it is a set of steps, with simple changes, you can check the intermediate results. Doing this, you will know where things do wrong.
-----
By the way, you gave an output, you forgot to say if it is actual output or expected output, and since you didn't gave the other output we don't know in what it is wrong.
Actual output is ...
Expected output is : ...
-----
Your code do not behave the way you expect, and you don't understand why !
There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.
Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
1.11 — Debugging your program (stepping and breakpoints) | Learn C++[^]
The debugger is here to only show you what your code is doing and your task is to compare with what it should do.
[Update]
Your example in Solution 3 is weird.
23
9 2 2 2 2 2 2 99 2 9 19
9 99 99 99 99 99 99 98 2 9 19
8 25 34 99 75 2 99 99 92 99
7 99 99 78 98 99 90 99 99
5 98 99 99 8 2 34
5 99 99 99 2 2 35
5 98 99 99 16 2 33
4 2 2 4 5 4
3 98 99 98 98
4 4 4 4 4 4
3 3 3 3 3
4 2 2 2 2 2
3 64 2 2 2
3 2 3 2 2
2 4 3 3
3 2 2 2 2
1 3 3
2 2 2 2
1 3 2
1 2 3
0 7
0 5
1 2 2
// correct answer
23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
//my answer
23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 1 2
// I would have expected
21 22 17 19 20 23 ...
因为
0 ^ x为0且1 ^ x为1
because
0 ^ x is 0 and 1 ^ x is 1
我的代码没有正确通过此测试。
输入
my code does not pass this test correctly.
input
23
9 2 2 2 2 2 2 99 2 9 19
9 99 99 99 99 99 99 98 2 9 19
8 25 34 99 75 2 99 99 92 99
7 99 99 78 98 99 90 99 99
5 98 99 99 8 2 34
5 99 99 99 2 2 35
5 98 99 99 16 2 33
4 2 2 4 5 4
3 98 99 98 98
4 4 4 4 4 4
3 3 3 3 3
4 2 2 2 2 2
3 64 2 2 2
3 2 3 2 2
2 4 3 3
3 2 2 2 2
1 3 3
2 2 2 2
1 3 2
1 2 3
0 7
0 5
1 2 2
// correct answer
23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
//my answer
23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 1 2
这篇关于比较10个塔。修复代码的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
