用 C++ 计算总和
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Calculating a Sum with C++
提问by Stripers247
I wrote the following code to sum the series (-1)^i*(i/(i+1)). But when I run it I get -1for any value of n.
我编写了以下代码来总结系列(-1)^i*(i/(i+1))。但是当我运行它时，我会得到-1任何 n 值。
Can some one please point out what I am doing wrong? Thank you in advance!
有人可以指出我做错了什么吗？先感谢您！
#include <iostream>
using namespace std;
int main()
{
int sum = 0;
int i = 1.0;
int n = 5.0;
for(i=1;i<=n;i++)
sum = (-1)^i*(i/(i+1));
cout << "Sum" <<" = "<< sum << endl;
return 0;
}
回答by Mysticial
Problem #1:The C++ ^operator isn't the math power operator. It's a bitwise XOR.
问题#1：C++^运算符不是数学幂运算符。这是一个按位异或。
You should use pow()instead.
你应该pow()改用。
Problem #2:You are storing floating-point types into an integer type. So the following will result in integer division (truncated division):
问题#2：您将浮点类型存储为整数类型。所以以下将导致整数除法（截断除法）：
i/(i+1)
Problem #3:You are not actually summing anything up:
问题 3：你实际上并没有总结任何东西：
sum = ...
should be:
应该：
sum += ...
A corrected version of the code is as follows:
代码的更正版本如下：
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++)
sum += pow(-1.,(double)i) * ((double)i / (i + 1));
Although you really don't need to use powin this case. A simple test for odd/even will do.
虽然你真的不需要pow在这种情况下使用。一个简单的奇数/偶数测试就可以了。
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++){
double val = (double)i / (i + 1);
if (i % 2 != 0){
val *= -1.;
}
sum += val;
}
回答by ClemPi
You need too put sum += pow(-1,i)*(i/(i+1));
你也需要 put sum += pow(-1,i)*(i/(i+1));
Otherwise you lose previous result each time.
否则你每次都会失去之前的结果。
Use pow function for pow operation.
使用 pow 函数进行 pow 操作。
edit : as said in other post, use double or float instead of int to avoid truncated division.
编辑：如其他帖子所述，使用 double 或 float 代替 int 以避免截断除法。
回答by Sergey Brunov
How about this
这个怎么样
((i % 2) == 0 ? 1 : -1)
instead of
代替
std::pow(-1, i)
?
?
Full answer:
完整答案：
double sum = 0;
int i = 1.0;
int n = 5.0;
for (i = 1; i <= n; ++i) {
signed char sign = ((i % 2) == 0 ? 1 : -1);
sum += sign * (i / (i+1));
}
回答by Rob?
You seem to have a few things wrong with your code:
您的代码似乎有一些问题：
using namespace std;
This is not directly related to your problem at hand, but don't ever say using namespace std;It introduces subtle bugs.
这与您手头的问题没有直接关系，但永远不要说它using namespace std;引入了微妙的错误。
int i = 1.0;
int n = 5.0;
You are initializaing integral variables with floating-point constants. Try
您正在使用浮点常量初始化整数变量。尝试
int i = 1;
int n = 5;
sum = (-1)^i*(i/(i+1));
You have two problems with this expression. First, the quantity (i/(i+1))is always zero. Remember dividing two ints rounds the result. Second, ^doesn't do what you think it does. It is the exclusive-or operator, not the exponentiation operator. Third, ^binds less tightly than *, so your expression is:
这个表达式有两个问题。首先，数量(i/(i+1))始终为零。记住除以两个ints 舍入结果。其次，^没有做你认为它会做的事情。它是异或运算符，而不是幂运算符。第三，^绑定不那么紧密*，所以你的表达式是：
-1 xor (i * (i/(i+1)))
-1 xor (i * 0)
-1 xor 0
-1
回答by Sid
Few problems:
几个问题：
^ is teh bitwise exclusive or in c++ not "raised to power". Use pow() method.
Remove the dangling opening bracket from the last line
Use ints not floats when assigning to ints.
^ 是按位排他的，或者在 C++ 中不是“提升到权力”。使用 pow() 方法。
从最后一行中删除悬空的开口支架
分配给整数时使用整数而不是浮点数。
回答by ssell
^does not do what you think it does. Also there are some other mistakes in your code.
^不会做你认为它会做的事情。您的代码中还有一些其他错误。
What it should be:
它应该是什么：
#include <iostream>
#include <cmath>
int main( )
{
long sum = 0;
int i = 1;
int n = 5;
for( i = 1; i <= n; i++ )
sum += std::pow( -1.f, i ) * ( i / ( i + 1 ) );
std::cout << "Sum = " << sum << std::endl;
return 0;
}
To take a power of a value, use std::pow(see here). Also you can not assign intto a decimal value. For that you need to use floator double.
要获取值的幂，请使用std::pow(请参见此处)。您也不能分配int给十进制值。为此，您需要使用float或double。
The aforementioned ^is a bitwise-XOR, not a mark for an exponent.
上述^是按位异或，而不是指数的标记。
Also be careful of Integer Arithmeticas you may get unexpected results. You most likely want to change your variables to either floator double.
还要小心整数运算，因为您可能会得到意想不到的结果。您很可能希望将变量更改为float或double。
回答by Andrei
There are a few issues with the code:
代码有几个问题：
int sum = 0;
The intermediate results are not integers, this should be a double
中间结果不是整数，应该是double
int i = 1.0;
Since you will use this in a division, it should be a double, 1/2 is 0 if calculated in integers.
由于您将在除法中使用它，因此它应该是双精度数，如果以整数计算，则 1/2 为 0。
int n = 5.0;
This is an int, not a floating point value, no .0 is needed.
这是一个整数，不是浮点值，不需要 .0。
for(i=1;i<=n;i++)
You've already initialized i to 1, why do it again?
你已经将 i 初始化为 1，为什么还要初始化？
sum = (-1)^i*(i/(i+1));
Every iteration you lose the previous value, you should use sum+= 'new values'
每次迭代都会丢失以前的值，您应该使用 sum+= 'new values'
Also, you don't need pow to calculate (-1)^i, all this does is switch between +1 and -1 depending on the odd/even status of i. You can do this easier with an ifstatement or with 2 for's, one for odd ione for even ones... Many choices really.
此外，您不需要 pow 来计算(-1)^i，所有这些都是根据 的奇数/偶数状态在 +1 和 -1 之间切换i。您可以使用if语句或 2更容易地做到这一点for，一个表示奇数，i一个表示偶数……确实有很多选择。
