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How do I print just a final sum in this problem?

I am a beginner, trying to solve Project Euler problem 1:
"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.", but, you probably know this problem.
So I get a correct answer but my program lists all sums up to a final sum. But how do I manage to just print final sum?
int x = 0;
for (int i = 1; i < 1000; i ++)
{
if (i % 3 == 0 || i % 5 == 0)
{
Console.WriteLine(x += i);

You have a Console.WriteLine inside a loop, so every time the if is true, a print out to the console will occur. Move the Console.WriteLine so it is outside of the loop
I think you would have quite quickly discovered this problem if you had used the debugger to step through the code line by line. Do you know how to use the debugger? If not, drop a comment and I'll write some introductory lines

You should keep console.WriteLine() outside of 'for' loop,
Your code should go look like this-
int x=0;
for (int i = 1; i < 1000; i ++){
if (i%3==0 || i%5==0){
x+=i;
}}
console.WriteLine(x);

good start bro, you can just make the Console.WriteLine(x) after the for loop

You don't need to write every time you add i to x. You can just add it and then print after you ran the for loop.
int x = 0;
for (int i = 0; i < 1000; i++)
{
if (i % 3 == 0 || i % 5 == 0)
{
x += i;
}
}
Console.WriteLine(x);

If you want to print only the final result, move the Console.WriteLine command out of the iteration and place it after it. This way it will execute only once.
int x = 0;
for (int i = 1; i < 1000; i ++)
{
if (i % 3 == 0 || i % 5 == 0)
{
x += i;
}
}
Console.WriteLine(x);

c# split string in 2. letter by letter

I need split my string in 2, one letter to each variable.
Example: string = "ABCDEFGHIJ"
name1: ACEGI
name2: BDFHJ
I done so far:
var builderM = new StringBuilder();
var builderK = new StringBuilder();
for (int i = 0; i < s.Length; i++)
{
builderM.Append(s[i]);
builderK.Append(s[i++]);
}
txtM.Text = builderM.ToString();
txtK.Text = builderK.ToString();
But its showing same text in the 2.

you should use ++i instead of i++
for (int i = 0; i < s.Length; i++)
{
builderM.Append(s[i]);
if(i + 1 < s.Length) // to prevent IOR exception when count is odd.
builderK.Append(s[++i]); // pre increment.
}
the reason is that i++ is post incremented. that means i gets incremented after the expression therefor s[i++] will give you same item as s[i].

Another approach would be to use LINQ to filter odd and even indices into two strings, something like:
var even = new string(input.Where((c, idx) => idx % 2 == 0).ToArray());
var odd = new string(input.Where((c, idx) => idx % 2 != 0).ToArray());

If performance is not an issue, you can use LINQ:
var name1 = String.Join(String.Empty, str.Where((v, i) => i % 2 == 0));
var name2 = String.Join(String.Empty, str.Where((v, i) => i % 2 == 1));

You can also use the modulus operator (%) to determine if the index is even or odd, and put the even indexes in the first array and the odd indexes in the second one:
for (int i = 0; i < s.Length; i++)
{
if (i % 2 == 0) builderM.Append(s[i]);
else builderK.Append(s[i]);
}
If you'd rather increment the i inside the for body, you have to repeat the check against s.Length (as we do in the for condition). Also, you will need to either move the post-increment to the previous line (so that i is incremented in time), or use a pre-increment:
// Move post-increment to previous line example:
for (int i = 0; i < s.Length; i++)
{
builderM.Append(s[i++]);
if (i < s.Length) builderK.Append(s[i]);
}
// Use a pre-increment example:
for (int i = 0; i < s.Length; i++)
{
builderM.Append(s[i]);
if (++i < s.Length) builderK.Append(s[i]);
}

max consecutive 1 or 0 that can be obtained by flipping one bit of binary nubmer

given a binary number find the maximum consecutive 1s or 0s that can be obtained by flipping only one bit (either a 1 or a 0)
the code given was
public int getMacCon(string[] A)
{
int n = A.Length;
int result = 0;
for (int i = 0; i < n - 1; i++)
{
if (A[i] == A[i + 1])
result = result + 1;
}
int r = -2;
for (int i = 0; i < n; i++)
{
int count = 0;
if (i > 0)
{
if (A[i - 1] != A[i])
count = count + 1;
else
count = count - 1;
}
if (i < n - 1)
{
if (A[i + 1] != A[i])
count = count + 1;
else
count = count - 1;
}
r = Math.Max(r, count);
}
return result + r;
}
Im finding hard to figure out the logic here. specially the given below part
for (int i = 0; i < n - 1; i++)
{
if (A[i] == A[i + 1])
result = result + 1;
}
I would highly apretiate if any one can explain me the logic in this solution.
Thanks

The bit you've highlighted just counts the number of currently adjacent equal values, i.e. where one value (A[i]) is equal to the next (A[i+1]). It then asks (the second loop), for each value in turn, whether flipping it would increase vs decrease vs not change the number of adjacent equal values; so if a value is currently different from the one before it (A[i-1] != A[i]), a flip would be an increase - else decrease; likewise the one after it. The final result is the pre-existing number of adjacent equal values (result), plus the best delta (r) found in the sweep.
public int getMacCon(string[] A)
{
int n = A.Length;
int result = 0;
// find how many adjacent values are currently in the string
for (int i = 0; i < n - 1; i++)
{
// if same as the next value, that's a hit
if (A[i] == A[i + 1])
result = result + 1;
}
// worst case delta from flipping one value is that me make it
// worse on both sides, so -2
int r = -2;
// now consider each value in turn
for (int i = 0; i < n; i++)
{
int count = 0;
if (i > 0) // test value before, if not the first value
{
if (A[i - 1] != A[i])
count = count + 1; // before is different; flip is incr
else
count = count - 1; // before is same; flip is decr
}
if (i < n - 1) // test value after, if not the last value
{
if (A[i + 1] != A[i])
count = count + 1; // after is different; flip is incr
else
count = count - 1; // after is same; flip is decr
}
// compare that to the tracking counter, and keep the best value
r = Math.Max(r, count);
}
// final result is sum of pre-existing count plus the delta
return result + r;
}
Incidentally, an optimization might be to change the second loop test from i < n to i < n && r != 2 - i.e. stop as soon as the best possible delta is found (making it better on both sides, +2)

Not an direct answer to your question (as Marc Gravell's answer covers it enough) but I just need to add how would I solve this instead:
encode your string with RLE (run length encoding)
so you need array of b,v,n values. Where b>=0 is start position, v={0,1} is bit value and n is the count of consequent occurencies. For example something like:
const int _max=100; // max number of bits
int b[_max],v[_max],n[_max],bvns=0;
The bvns is number of used b,v,n in the arrays. You can also use any dynamic list/template instead.
reset your actual solution
You need bit position to change ix and the count of consequent bits resulting after its flip sz. Set booth to zero.
scan the RLE for items with n=1
if found that means item before and after in the RLE is the same so flipping will join them. So compute the resulting size and if bigger then store as actual solution something like:
for (int i=1;i<bvns-1;i++) // scann RLE except first and last item
if (n[i]==1) // single bit found
{
l=n[i-1]+1+n[i+1]; // resulting size
if (l>=sz) { sz=l; ix=b; } // update solution
}
test if enlarging single sequence is not bigger
simply:
for (int i=0;i<bvns;i++) // scann RLE
if (n[i]>=sz) // result is bigger?
{
sz=l; ix=b-1; // update solution
if (ix<0) ix=b+n[i];
}

Need Help using the Mod Function for looping in C#

for (int i = 0; i < 100; i = (i%10))
{
age = (age +10);
int_year = (int_year + 10);
So I am trying to use the Mod function for looping in C#, the requirement is that it the int_year and age loop 100 times but only every 10th increment is kept. Thanks in advance :)

Why not just mod 10 within the loop to determine if this is the 10th iteration:
for (int i = 0; i < 100; i++)
{
if (i % 10 == 0)
{
// Every 10th iteration
}
}

You can use step equal to 10:
for (int i = 0; i < 100; i += 10)
{
// Every 10th iteration
}
I think it's more efficient than using mod.

Checkers Board Assistance

I'm just wondering if there is a simpler way of doing this:
for (int i = 0; i < 1; i++)
{
for (int j = 0; i < 8; j+2)
{
board[ i, j ] = 2;
board[( i + 1 ), j ] = 2;
board[( i + 2 ), j ] = 2;
}
}
What I'm trying to do is place checkers pieces on the actual checkers board. So this is to place the black pieces on the top.
P.S. If you could also give me some help on the bottom set of pieces(white).

Apart from fixing the loop you could otherwise explicitly place the pieces, makes it more readable
int[,] board = new[,]{{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1},
{1,0,1,0,1,0,1,0},
{0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0},
{0,1,0,1,0,1,0,1},
{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1}};

Modulo will do the trick but i think that TonyS's anwser was my first reaction and I would prefer that insteed the one shown below.
char[,] board = new char[8,8];
private void InitializeBoard()
{
string BoardRepresentation = "";
//for every board cell
for (int i = 0; i < 8; i++)
{
for (int j = 0; j < 8; j++)
{
//initialize board cell
board[i, j] = '0';
if (j <= 2)//black top
{
//Modulo is the trick
if ((j - i) == 0 || ((j - i) % 2) == 0)
{
board[i, j] = 'B';
}
}
else if (j >= 5) //white bot
{
if ((j - i) == 0 || ((j - i) % 2) == 0)
{
board[i, j] = 'W';
}
}
}
}
for (int j = 0; j < 8; j++)
{
for (int i = 0; i < 8; i++)
{
BoardRepresentation += board[i, j] + " ";
}
BoardRepresentation += Environment.NewLine;
}
}

You missed to tell us what exactly is it that you want to achieve. I see several big errors in your code, so I'll assume that you don't have full grasp of how for loop works. I hope it's not too presumptuous that I'm trying to explain it here
For loop
For loop is used to execute same portion of code several times. How many times it will be executed depends on conditions you set. Most often, you will see it in this format:
for (int i = 0; i < n; i++)
{
// Some code
}
This for loop executed code within the brackets ({ and }) n times. This is not only way to define a loop. More thorough definition of a loop is following:
for (<initialization>; <condition>; <afterthought>)
Initialization - You can some variables needed for looping. This is executed once before code within the loop is executed first time. This is optional, and you can leave it empty and use variable declared before in condition.
Condition - This is executed before each execution of code within the loop. If condition expression evaluates to true, loop is executed. Once loop is executed and afterthought is executed, condition is evaluated again and again until it evaluates to false. Condition is also optional. If you leave it out, in C# loop will be executed again until you break the loop in a different way.
Afterthought - This is executed each time code within the loop is finished executing. This is usually used to increment a variable on which loop depends. This is also optional.
Fixing your code
I assume you wanted to mark fields in a 8x8 matrix like in a checkerboard, although this is not stated in your question. You could do it this way:
// For each row in a board (start with 0, go until 7, increase by 1)
for (int i = 0; i < 8; i++)
{
// start coloring the row. Determine which field within the row needs
// to be black. In first row, first field is black, in second second
// field is black, in third row first field is black again and so on.
// So, even rows have black field in first blace, odd rows have black
// on second place.
// We calculate this by determining division remained when dividing by 2.
int firstBlack = i % 2;
// Starting with calculated field, and until last field color fields black.
// Skip every second field. (start with firstBlack, go until 7, increase by 2)
for (int j = firstBlack; j < 8; j += 2)
{
// Color the field black (set to 2)
board[i][j] = 2;
}
}
You can see my comments inline.
Big errors in your code
// This loop would be executed only once. It goes from 0 to less than 1 and is increased
// after first execution. You might as well done it without the loop.
for (int i = 0; i < 1; i++)
{
// This doesn't make sense, because you use i in condition, and initialize
// and update j.
// Also, you are trying to update j, but you are not doing so. You are not
// assigning anything to you. You should do j+=2 to increase by two. Or you
// can do j = j + 2
for (int j = 0; i < 8; j+2)
{
// I didn't really understand what you were trying to achieve here
board[ i, j ] = 2;
board[( i + 1 ), j ] = 2;
board[( i + 2 ), j ] = 2;
}
}