C程序检查是否有一个等于或等于零的子数组 [英] C program to check if there is a an subarray of sum equal to zero or not
问题描述
我做了一个ac程序来检查是否有一个零和零的子数组
这里的约束是我必须使用循环这个问题这是我试过的程序
i我是编程的新手
我尝试了什么：
i made a c program to check if there is a sub array with sum zero or not
the constraint here is that i have to do this problem using loop only this is my program that i have tried
i am new to programming
What I have tried:
#include<stdio.h>
#include<stdlib.h>
int checksum(int* a,int size);
int main()
{
int size,i,flag=0;
printf("enter the size of array");
scanf("%d",&size);
int* a=(int*)malloc(size*sizeof(int));
for(i=0;i<size;i++)
scanf("%d",&a[i]);
printf("the entered array is\n ");
for(i=0;i<size;i++)
printf("%d\n",*(a+i));
flag=checksum(a,size);
if(flag==1)
printf("subarray with sum 0 is present");
else
printf("subarray with sum zero not present");
}
这是我检查金额的功能
this is my function for checking the sum
int checksum(int* a,int size)
{ int i,k,sum=0,flag;
for(i=0;i<size;i++){
sum=a[i];
for(k=0;k<size;k++)
{
sum=sum+a[k];
if(sum==0){
flag=1;
break;}
}
}
return flag;
}
i我得到正确的输出bt我对我的程序不满意任何人都可以帮我纠正这个程序
note-use循环，使用指针，至少有一个函数而不是main
i am getting the correct output bt i am not satisfied with my program can anyone help me correcting this program
note-use loop only,use of pointers,at least one function other than main
推荐答案
好的，所以现在你已经编写了一些代码 - 或者至少抛出了一些人在一起首先重构你的代码，使其缩进和可读：使用有意义的名称而不是你可以逃脱的最短名称，并选择包围样式并保持一致。如果你使用K& R，Whitesmiths，甚至是可执行的1TB都没关系，一致的缩进使你的代码更具可读性！
OK, so now you have written some code - or at least thrown some together. Start by refactoring you code so it's indented and readable: use meaningful names instead of the shortest you can get away with, and pick an bracketing style and be consistent. It doesn't matter if you use K&R, Whitesmiths, or even the execrable 1TB, consistent indentation makes your code so much more readable!
int checksum(int* a,int size)
{
int i,k,sum=0,flag;
for(i=0;i<size;i++)
{
sum=a[i];
for(k=0;k<size;k++)
{
sum=sum+a[k];
if(sum==0)
{
flag=1;
break;
}
}
}
return flag;
}现在你可以一眼看出你的功能控制流程是什么。
现在它是是时候进入下一步了。编译并不意味着您的代码是正确的！ ：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：
Now you can see at a glance what the flow of control through your function is going to be.
Now it's time to move to the next step. 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。我无法确切地告诉您要使用调试器的按键是什么 - 有多种不同的方式与编译器一样 - 但是快速的Google用于IDE的名称和调试器应该为您提供所需的信息。 br />
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. I can't tell you exactly what keys to press to use the debugger - there are as many different ways as there are compilers - but a quick Google for the name of your IDE and "debugger" should give you the info you need.
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!
你的函数有问题，你没有在开始时初始化flag。 br />
Your function have a problem, you do not initializeflagat begining.
int i,k,sum=0,flag=0;
可能带来一些误报。<学习如何正确缩进代码，显示其结构，有助于阅读和理解。\\ n \\ n
它还有助于发现结构错误。
which can bring some false positive.
Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.
int checksum(int* a,int size){
int i,k,sum=0,flag;
for(i=0;i<size;i++){
sum=a[i];
for(k=0;k<size;k++)
{
sum=sum+a[k];
if(sum==0){
flag=1;
break;
}
}
}
return flag;
}
专业程序员的编辑器具有此功能，其他功能包括括号匹配和语法高亮。
Notepad++主页 [ ^ ]
ultraedit [ ^ ]
[更新]
By顺便说一下，在二读时，你的函数代码很奇怪，检查你的总和。
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