错误简单输出C ++ [英] Error simple output C++

问题描述

你好帮助简单的输出程序..

当int非常大是错误的时候..

还有更好的解决方案吗？

喜欢

n = 1000000000000

k = 1000000000001

这将是错误



我尝试过的事情：

hello help simple output program..
when the int very large is error..
is there a better solution?
like
n = 1000000000000
k = 1000000000001
this will be error

What I have tried:

long long n;
  long long k;
	unsigned int count = 0;
	cin >> n >> k;
	for(int i = 1; i <= n; i++)
	{
		for(int y = i+1; y <= n; y++)
		{
			if(i + y == k)
			{
				count = count + 1;
			}
		}
	}
	cout << count << endl;
	return 0;

推荐答案

在大多数平台上，C / C ++类型 unsigned int 是32位宽。这意味着最大值。十进制值为4,294,967,295，小于1,000,000,000,000。



如果需要更大的值，请使用64位值： unsigned long long 或 uint64_t （包括 cstdint 与C ++ 11或 stdint.h ）。

On most platforms the C/C++ type unsigned int is 32-bit wide. That means the max. decimal value is 4,294,967,295 which is less than 1,000,000,000,000.

If you need larger values use 64 bit values: unsigned long long or uint64_t (include cstdint with C++11 or stdint.h).

你对n和k使用long long类型，它也需要用于i，y和k，因为正如Jochen写的那样，在达到n和n之前将达到计数限制k。

You used the long long type for n and k and it also needs to be used for i, y, and k because, as Jochen wrote, the limit for count will be reached before it is reached for n and k.

你知道有不同类型的整数，大小不同。

你需要保持一致，你不能有 n 和 k 作为long long类型并期望使用int作为计数器，其中n为循环限制。



2个嵌套循环非常简单，因此非常粗暴。

拿一张纸和一支铅笔用小值解决问题，编写每个解决方案或更改你的代码，为每个解决方案打印i和j。

做一点分析，你应该很容易删除内循环。

进一步分析，你可以删除两个循环并找到一个数学公式，直接给你答案。

You know that there is different types of integers, different sizes.
You need to be consistent, you can't have n and k as long long type and expect to use ints as counters with n as the loop limit.

The 2 nested loops are very simple minded, and thus brute force.
Take a sheet of paper and a pencil and solve the problem with small values, write each solution or change your code to print i and j for each solution.
do a little analyze, you should easily be able to remove the inner loop.
with further analyze, you may be able to remove both loops and find a mathematical formula which will give you directly the answer.

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