I've been working on a program for my Algorithm Analysis class where I have to solve the Knapsack problem with Brute Force, greedy, dynamic, and branch and bound strategies. Everything works perfectly when I run it in Visual Studio 2012, but if I compile with gcc and run it on the command line, I get a different result:
Visual Studio:
+-------------------------------------------------------------------------------+
| Number of | Processing time in seconds / Maximum benefit value |
| +---------------+---------------+---------------+---------------+
| items | Brute force | Greedy | D.P. | B. & B. |
+---------------+---------------+---------------+---------------+---------------+
| 10 + 0 / 1290 + 0 / 1328 + 0 / 1290 + 0 / 1290 |
+---------------+---------------+---------------+---------------+---------------+
| 20 + 0 / 3286 + 0 / 3295 + 0 / 3200 + 0 / 3286 |
+---------------+---------------+---------------+---------------+---------------+
cmd:
+-------------------------------------------------------------------------------+
| Number of | Processing time in seconds / Maximum benefit value |
| +---------------+---------------+---------------+---------------+
| items | Brute force | Greedy | D.P. | B. & B. |
+---------------+---------------+---------------+---------------+---------------+
| 10 + 0 / 1290 + 0 / 1328 + 0 / 1599229779+ 0 / 1290 |
+---------------+---------------+---------------+---------------+---------------+
| 20 + 0 / 3286 + 0 / 3295 + 0 / 3200 + 0 / 3286 |
+---------------+---------------+---------------+---------------+---------------+
The same number always shows up, "1599229779." Notice that the output is only messed up the first time the Dynamic algorithm is run.
Here is my code:
typedef struct{
short value; //This is the value of the item
short weight; //This is the weight of the item
float ratio; //This is the ratio of value/weight
} itemType;
typedef struct{
time_t startingTime;
time_t endingTime;
int maxValue;
} result;
result solveWithDynamic(itemType items[], int itemsLength, int maxCapacity){
result answer;
int rowSize = 2;
int colSize = maxCapacity + 1;
int i, j; //used in loops
int otherColumn, thisColumn;
answer.startingTime = time(NULL);
int **table = (int**)malloc((sizeof *table) * rowSize);//[2][(MAX_ITEMS*WEIGHT_MULTIPLIER)];
for(i = 0; i < rowSize; i ++)
table[i] = (int*)malloc((sizeof *table[i]) * colSize);
table[0][0] = 0;
table[1][0] = 0;
for(i = 1; i < maxCapacity; i++) table[1][i] = 0;
for(i = 0; i < itemsLength; i++){
thisColumn = i%2;
otherColumn = (i+1)%2; //this is always the other column
for(j = 1; j < maxCapacity + 1; j++){
if(items[i].weight <= j){
if(items[i].value + table[otherColumn][j-items[i].weight] > table[otherColumn][j])
table[thisColumn][j] = items[i].value + table[otherColumn][j-items[i].weight];
else
table[thisColumn][j] = table[otherColumn][j];
} else {
table[thisColumn][j] = table[thisColumn][j-1];
}//end if/else
}//end for
}//end for
answer.maxValue = table[thisColumn][maxCapacity];
answer.endingTime = time(NULL);
for(i = 0; i < rowSize; i ++)
free(table[i]);
free(table);
return answer;
}//end solveWithDynamic
Just a bit of explanation. I was having trouble with the memory consumption of this algorithm because I have to run it for a set of 10,000 items. I realized that I didn't need to store the whole table, because I only ever looked at the previous column. I actually figured out that you only need to store the current row and x+1 additional values, where x is the weight of the current itemType. It brought the memory required from (itemsLength+1) * (maxCapacity+1) elements to 2*(maxCapacity+1) and possibly (maxCapacity+1) + (x+1) (although I don't need to optimize it that much).
Also, I used printf("%d", answer.maxValue); in this function, and it still came out as "1599229779." Can anyone help me figure out what is going on? Thanks.
Can't be sure that that is what causes it, but
for(i = 1; i < maxCapacity; i++) table[1][i] = 0;
you leave table[1][maxCapacity] uninitialised, but then potentially use it:
for(j = 1; j < maxCapacity + 1; j++){
if(items[i].weight <= j){
if(items[i].value + table[otherColumn][j-items[i].weight] > table[otherColumn][j])
table[thisColumn][j] = items[i].value + table[otherColumn][j-items[i].weight];
else
table[thisColumn][j] = table[otherColumn][j];
} else {
table[thisColumn][j] = table[thisColumn][j-1];
}//end if/else
}//end for
If that is always zero with Visual Studio, but nonzero with gcc, that could explain the difference.
Related
Array A contains the elements, A1,A2...AN. And array B contains the elements, B1,B2...BN. There is a relationship between Ai and Bi, for 1 = i = N, i.e.,
any element Ai lies between 1 and Bi.
Let the cost S of an array A be defined as:
You have to print the largest possible value of S.
The Link to the problem is Problem
Example:
size of array:5
array: 10 1 10 1 10
output : 36 (since the max value can be derived as |10 - 1| + |1 - 10| + |10 - 1| + |1 - 10|)
Approach :
The only approach i could think of was brute force. I thought i would make a overlapping recursive equation so that i could memoize it, but was not able to.
CODE :
public static void func(int pos,int[] arr,int[] aux,int n)
{
/*
* pos is current index in the arr
* arr is array
* aux is temp array which will store one possible combination.
* n is size of the array.
* */
//if reached at the end, check the summation of differences
if(pos == n)
{
long sum = 0;
for(int i = 1 ; i < n ; i++)
{
//System.out.print("i = " + i + ", arr[i] = " + aux[i] + " ");
sum += Math.abs(aux[i] - aux[i - 1]);
}
//System.out.println();
//System.out.println("sum = " + sum);
if(sum > max)
{
max = sum;
}
return;
}
//else try every combination possible.
for(int i = 1 ; i <= arr[pos] ; i++)
{
aux[pos] = i;
func(pos + 1,arr,aux,n);
}
}
NOTE:
The complexity of this is O(n*2^n)
First, there is no reason that a[i] should be equal to any number besides 1 and b[i]. Realizing that we can write down a simple recurrence:
fmax(1) = fone(1) = 0
fmax(i) = max(fone(i-1) + b[i] - 1, fmax(i-1) + abs(b[i]-b[i-1]))
fone(i) = max(fone(i-1), fmax(i-1) + b[i-1] - 1)
answer = max(fmax(N), fone(N))
Where fmax(i) is a maximal sum for a[1..i] elements that end with b[i], fone(i) is a maximal sum for a[1..i] elements that end with 1.
With dynamic programming approach, the complexity is O(N).
Here is the code I am using. When I run it, it doesn't seem to change anything in the image except the last 1/4 of it. That part turns to a solid color.
void maxFilter(pixel * data, int w, int h)
{
GLubyte tempRed;
GLubyte tempGreen;
GLubyte tempBlue;
int i;
int j;
int k;
int pnum = 0;
int pnumWrite = 0;
for(i = 0 ; i < (h - 2); i+=3) {
for(j = 0 ; j < (w - 2); j+=3) {
tempRed = 0;
tempGreen = 0;
tempBlue = 0;
for (k = 0 ; k < 3 ; k++){
if ((data[pnum].r) > tempRed){tempRed = (data[pnum + k].r);}
if ((data[pnum].g) > tempGreen){tempGreen = (data[pnum + k].g);}
if ((data[pnum].b) > tempBlue){tempBlue = (data[pnum + k].b);}
if ((data[(pnum + w)].r) > tempRed){tempRed = (data[(pnum + w)].r);}
if ((data[(pnum + w)].g) > tempGreen){tempGreen = (data[(pnum + w)].g);}
if ((data[(pnum + w)].b) > tempBlue){tempBlue = (data[(pnum + w)].b);}
if ((data[(pnum + 2 * w)].r) > tempRed){tempRed = (data[(pnum + 2 * w)].r);}
if ((data[(pnum + 2 * w)].g) > tempGreen){tempGreen = (data[(pnum + 2 * w)].g);}
if ((data[(pnum + 2 * w)].b) > tempBlue){tempBlue = (data[(pnum + 2 * w)].b);}
pnum++;
}
pnumWrite = pnum - 3;
for (k = 0 ; k < 3 ; k++){
((data[pnumWrite].r) = tempRed);
((data[pnumWrite].g) = tempGreen);
((data[pnumWrite].b) = tempBlue);
((data[(pnumWrite + w)].r) = tempRed);
((data[(pnumWrite + w)].g) = tempGreen);
((data[(pnumWrite + w)].b) = tempBlue);
((data[(pnumWrite + 2 * w)].r) = tempRed);
((data[(pnumWrite + 2 * w)].g) = tempGreen);
((data[(pnumWrite + 2 * w)].b) = tempBlue);
pnumWrite++;
}
}
}
}
I can see several problems with that code - being difficult to follow not being the least!
I think your main problem is that the loop is (as you probably intended) run through h/3 * w/3 times, once for each 3x3 block in the image. But the pnum index runs only increases by 3 for each block, and reaches a maximum of about h*w/3, rather than the intended h*w. That means that only the first third of your image will be affected by your filter. (And I suspect your painting is done 'bottom-up', so that's why you see the lowest part change. I remember .bmp files being structured that way, but perhaps there are others as well.)
The 'cheap' fix would be to add 2*w at the right point, but nobody will ever understand that code again. I suggest you rewrite your indexing instead, and explicitly compute pnum from i and j in each turn through the loop. That can be improved on for readability, but is reasonably clear.
There's another minor thing: you have code like
if ((data[pnum].r) > tempRed){tempRed = (data[pnum + k].r);}
where the indexing on the right and on the left differ: this is probably also giving you results different from what you intended.
As Jongware points out, writing to the input array is always dangerous - your code is intended, I believe, to avoid that problem by only looking once into each 3x3 block, but his suggestion of a separate output array is very sensible - you probably don't want the blockiness your code gives anyway (you make each 3x3 block all one colour, don't you?), and his suggestion would let you avoid that.
Recently I wrote a program in C. During code execution, data calculation is bottleneck. As following:
The data structure is:
typedef struct tuple_t{
int oid;
int min_x;
int min_y;
int max_x;
int max_y;
}tuple_t
the code is
for (i = 0; i < Qry->num_tuples; i++) {
tuple_t Qi = Qry->tuples[i];
for (j = 0; j < Obj->num_tuples; j++) {
tuple_t Oj = Obj->tuples[j];
int test_top_bit = (Oj.min_x - Qi.min_x) | (Qi.max_x - Oj.min_x)
| (Oj.min_y - Qi.min_y) | (Qi.max_y - Oj.min_y);
test_top_bit >= 0 ? matches++ : 0;
}
}
The code is uesd for testing whether a point is in a rectangle in 2 dimension.
The Qry->num_tuples and Obj->num_tuple is 5 million. I run the test, the time is 887 millionseconds.
And I test the clasue
if(Oj.min_x == Qi.min_x)
count++;
the time is only 3 millionseconds. So the major time is spent on the clause:
int test_top_bit = (Oj.min_x - Qi.min_x) | (Qi.max_x - Oj.min_x)
| (Oj.min_y - Qi.min_y) | (Qi.max_y - Oj.min_y);
test_top_bit >= 0 ? matches++ : 0;
I used another join algorithms, but the time is still very long.
Is there anyways to improve the performance of the testing?Could SSE of SIMD be using ?
Looking at this line I see a performance problem:
tuple_t Oj = Obj->tuples[j];
You copy this struct 25 trillion times for no reason except cleaner code.
Try using a pointer instead.
tuple_t* pOj = &Obj->tuples[j];
You can also avoid branch:
matches += ( (Oj.min_x - Qi.min_x) | (Qi.max_x - Oj.min_x) |
(Oj.min_y - Qi.min_y) | (Qi.max_y - Oj.min_y) ) >=0;
Can anyone spot any way to improve the speed in the next Bilinear resizing Algorithm?
I need to improve Speed as this is critical, keeping good image quality. Is expected to be used in mobile devices with low speed CPUs.
The algorithm is used mainly for up-scale resizing. Any other faster Bilinear algorithm also would be appreciated. Thanks
void resize(int* input, int* output, int sourceWidth, int sourceHeight, int targetWidth, int targetHeight)
{
int a, b, c, d, x, y, index;
float x_ratio = ((float)(sourceWidth - 1)) / targetWidth;
float y_ratio = ((float)(sourceHeight - 1)) / targetHeight;
float x_diff, y_diff, blue, red, green ;
int offset = 0 ;
for (int i = 0; i < targetHeight; i++)
{
for (int j = 0; j < targetWidth; j++)
{
x = (int)(x_ratio * j) ;
y = (int)(y_ratio * i) ;
x_diff = (x_ratio * j) - x ;
y_diff = (y_ratio * i) - y ;
index = (y * sourceWidth + x) ;
a = input[index] ;
b = input[index + 1] ;
c = input[index + sourceWidth] ;
d = input[index + sourceWidth + 1] ;
// blue element
blue = (a&0xff)*(1-x_diff)*(1-y_diff) + (b&0xff)*(x_diff)*(1-y_diff) +
(c&0xff)*(y_diff)*(1-x_diff) + (d&0xff)*(x_diff*y_diff);
// green element
green = ((a>>8)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>8)&0xff)*(x_diff)*(1-y_diff) +
((c>>8)&0xff)*(y_diff)*(1-x_diff) + ((d>>8)&0xff)*(x_diff*y_diff);
// red element
red = ((a>>16)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>16)&0xff)*(x_diff)*(1-y_diff) +
((c>>16)&0xff)*(y_diff)*(1-x_diff) + ((d>>16)&0xff)*(x_diff*y_diff);
output [offset++] =
0x000000ff | // alpha
((((int)red) << 24)&0xff0000) |
((((int)green) << 16)&0xff00) |
((((int)blue) << 8)&0xff00);
}
}
}
Off the the top of my head:
Stop using floating-point, unless you're certain your target CPU has it in hardware with good performance.
Make sure memory accesses are cache-optimized, i.e. clumped together.
Use the fastest data types possible. Sometimes this means smallest, sometimes it means "most native, requiring least overhead".
Investigate if signed/unsigned for integer operations have performance costs on your platform.
Investigate if look-up tables rather than computations gain you anything (but these can blow the caches, so be careful).
And, of course, do lots of profiling and measurements.
In-Line Cache and Lookup Tables
Cache your computations in your algorithm.
Avoid duplicate computations (like (1-y_diff) or (x_ratio * j))
Go through all the lines of your algorithm, and try to identify patterns of repetitions. Extract these to local variables. And possibly extract to functions, if they are short enough to be inlined, to make things more readable.
Use a lookup-table
It's quite likely that, if you can spare some memory, you can implement a "store" for your RGB values and simply "fetch" them based on the inputs that produced them. Maybe you don't need to store all of them, but you could experiment and see if some come back often. Alternatively, you could "fudge" your colors and thus end up with less values to store for more lookup inputs.
If you know the boundaries for you inputs, you can calculate the complete domain space and figure out what makes sense to cache. For instance, if you can't cache the whole R, G, B values, maybe you can at least pre-compute the shiftings ((b>>16) and so forth...) that are most likely deterministic in your case).
Use the Right Data Types for Performance
If you can avoid double and float variables, use int. On most architectures, int would be test faster type for computations because of the memory model. You can still achieve decent precision by simply shifting your units (ie use 1026 as int instead of 1.026 as double or float). It's quite likely that this trick would be enough for you.
x = (int)(x_ratio * j) ;
y = (int)(y_ratio * i) ;
x_diff = (x_ratio * j) - x ;
y_diff = (y_ratio * i) - y ;
index = (y * sourceWidth + x) ;
Could surely use some optimization: you were using x_ration * j-1 just a few cycles earlier, so all you really need here is x+=x_ratio
My random guess (use a profiler instead of letting people guess!):
The compiler has to generate that works when input and output overlap which means it has to do generate loads of redundant stores and loads. Add restrict to the input and output parameters to remove that safety feature.
You could also try using a=b; and c=d; instead of loading them again.
here is my version, steal some ideas. My C-fu is quite weak, so some lines are pseudocodes, but you can fix them.
void resize(int* input, int* output,
int sourceWidth, int sourceHeight,
int targetWidth, int targetHeight
) {
// Let's create some lookup tables!
// you can move them into 2-dimensional arrays to
// group together values used at the same time to help processor cache
int sx[0..targetWidth ]; // target->source X lookup
int sy[0..targetHeight]; // target->source Y lookup
int mx[0..targetWidth ]; // left pixel's multiplier
int my[0..targetHeight]; // bottom pixel's multiplier
// we don't have to calc indexes every time, find out when
bool reloadPixels[0..targetWidth ];
bool shiftPixels[0..targetWidth ];
int shiftReloadPixels[0..targetWidth ]; // can be combined if necessary
int v; // temporary value
for (int j = 0; j < targetWidth; j++){
// (8bit + targetBits + sourceBits) should be < max int
v = 256 * j * (sourceWidth-1) / (targetWidth-1);
sx[j] = v / 256;
mx[j] = v % 256;
reloadPixels[j] = j ? ( sx[j-1] != sx[j] ? 1 : 0)
: 1; // always load first pixel
// if no reload -> then no shift too
shiftPixels[j] = j ? ( sx[j-1]+1 = sx[j] ? 2 : 0)
: 0; // nothing to shift at first pixel
shiftReloadPixels[j] = reloadPixels[i] | shiftPixels[j];
}
for (int i = 0; i < targetHeight; i++){
v = 256 * i * (sourceHeight-1) / (targetHeight-1);
sy[i] = v / 256;
my[i] = v % 256;
}
int shiftReload;
int srcIndex;
int srcRowIndex;
int offset = 0;
int lm, rm, tm, bm; // left / right / top / bottom multipliers
int a, b, c, d;
for (int i = 0; i < targetHeight; i++){
srcRowIndex = sy[ i ] * sourceWidth;
tm = my[i];
bm = 255 - tm;
for (int j = 0; j < targetWidth; j++){
// too much ifs can be too slow, measure.
// always true for first pixel in a row
if( shiftReload = shiftReloadPixels[ j ] ){
srcIndex = srcRowIndex + sx[j];
if( shiftReload & 2 ){
a = b;
c = d;
}else{
a = input[ srcIndex ];
c = input[ srcIndex + sourceWidth ];
}
b = input[ srcIndex + 1 ];
d = input[ srcIndex + 1 + sourceWidth ];
}
lm = mx[j];
rm = 255 - lm;
// WTF?
// Input AA RR GG BB
// Output RR GG BB AA
if( j ){
leftOutput = rightOutput ^ 0xFFFFFF00;
}else{
leftOutput =
// blue element
((( ( (a&0xFF)*tm
+ (c&0xFF)*bm )*lm
) & 0xFF0000 ) >> 8)
// green element
| ((( ( ((a>>8)&0xFF)*tm
+ ((c>>8)&0xFF)*bm )*lm
) & 0xFF0000 )) // no need to shift
// red element
| ((( ( ((a>>16)&0xFF)*tm
+ ((c>>16)&0xFF)*bm )*lm
) & 0xFF0000 ) << 8 )
;
}
rightOutput =
// blue element
((( ( (b&0xFF)*tm
+ (d&0xFF)*bm )*lm
) & 0xFF0000 ) >> 8)
// green element
| ((( ( ((b>>8)&0xFF)*tm
+ ((d>>8)&0xFF)*bm )*lm
) & 0xFF0000 )) // no need to shift
// red element
| ((( ( ((b>>16)&0xFF)*tm
+ ((d>>16)&0xFF)*bm )*lm
) & 0xFF0000 ) << 8 )
;
output[offset++] =
// alpha
0x000000ff
| leftOutput
| rightOutput
;
}
}
}
I have had to start to learning C as part of a project that I am doing. I have started doing the 'euler' problems in it and am having trouble with the first one. I have to find the sum of all multiples of 3 or 5 below 1000. Could someone please help me. Thanks.
#include<stdio.h>
int start;
int sum;
int main() {
while (start < 1001) {
if (start % 3 == 0) {
sum = sum + start;
start += 1;
} else {
start += 1;
}
if (start % 5 == 0) {
sum = sum + start;
start += 1;
} else {
start += 1;
}
printf("%d\n", sum);
}
return(0);
}
You've gotten some great answers so far, mainly suggesting something like:
#include <stdio.h>
int main(int argc, char * argv[])
{
int i;
int soln = 0;
for (i = 1; i < 1000; i++)
{
if ((i % 3 == 0) || (i % 5 == 0))
{
soln += i;
}
}
printf("%d\n", soln);
return 0;
}
So I'm going to take a different tack. I know you're doing this to learn C, so this may be a bit of a tangent.
Really, you're making the computer work too hard for this :). If we figured some things out ahead of time, it could make the task easier.
Well, how many multiples of 3 are less than 1000? There's one for each time that 3 goes into 1000 - 1.
mult3 = ⌊ (1000 - 1) / 3 ⌋ = 333
(the ⌊ and ⌋ mean that this is floor division, or, in programming terms, integer division, where the remainder is dropped).
And how many multiples of 5 are less than 1000?
mult5 = ⌊ (1000 - 1) / 5 ⌋ = 199
Now what is the sum of all the multiples of 3 less than 1000?
sum3 = 3 + 6 + 9 + ... + 996 + 999 = 3×(1 + 2 + 3 + ... + 332 + 333) = 3×∑i=1 to mult3 i
And the sum of all the multiples of 5 less than 1000?
sum5 = 5 + 10 + 15 + ... + 990 + 995 = 5×(1 + 2 + 3 + ... + 198 + 199) = 5×∑i = 1 to mult5 i
Some multiples of 3 are also multiples of 5. Those are the multiples of 15.
Since those count towards mult3 and mult5 (and therefore sum3 and sum5) we need to know mult15 and sum15 to avoid counting them twice.
mult15 = ⌊ (1000 - 1) /15 ⌋ = 66
sum15 = 15 + 30 + 45 + ... + 975 + 990 = 15×(1 + 2 + 3 + ... + 65 + 66) = 15×∑i = 1 to mult15 i
So the solution to the problem "find the sum of all the multiples of 3 or 5 below 1000" is then
soln = sum3 + sum5 - sum15
So, if we wanted to, we could implement this directly:
#include <stdio.h>
int main(int argc, char * argv[])
{
int i;
int const mult3 = (1000 - 1) / 3;
int const mult5 = (1000 - 1) / 5;
int const mult15 = (1000 - 1) / 15;
int sum3 = 0;
int sum5 = 0;
int sum15 = 0;
int soln;
for (i = 1; i <= mult3; i++) { sum3 += 3*i; }
for (i = 1; i <= mult5; i++) { sum5 += 5*i; }
for (i = 1; i <= mult15; i++) { sum15 += 15*i; }
soln = sum3 + sum5 - sum15;
printf("%d\n", soln);
return 0;
}
But we can do better. For calculating individual sums, we have Gauss's identity which says the sum from 1 to n (aka ∑i = 1 to n i) is n×(n+1)/2, so:
sum3 = 3×mult3×(mult3+1) / 2
sum5 = 5×mult5×(mult5+1) / 2
sum15 = 15×mult15×(mult15+1) / 2
(Note that we can use normal division or integer division here - it doesn't matter since one of n or n+1 must be divisible by 2)
Now this is kind of neat, since it means we can find the solution without using a loop:
#include <stdio.h>
int main(int argc, char *argv[])
{
int const mult3 = (1000 - 1) / 3;
int const mult5 = (1000 - 1) / 5;
int const mult15 = (1000 - 1) / 15;
int const sum3 = (3 * mult3 * (mult3 + 1)) / 2;
int const sum5 = (5 * mult5 * (mult5 + 1)) / 2;
int const sum15 = (15 * mult15 * (mult15 + 1)) / 2;
int const soln = sum3 + sum5 - sum15;
printf("%d\n", soln);
return 0;
}
Of course, since we've gone this far we could crank out the entire thing by hand:
sum3 = 3×333×(333+1) / 2 = 999×334 / 2 = 999×117 = 117000 - 117 = 116883
sum5 = 5×199×(199+1) / 2 = 995×200 / 2 = 995×100 = 99500
sum15 = 15×66×(66+1) / 2 = 990×67 / 2 = 495 × 67 = 33165
soln = 116883 + 99500 - 33165 = 233168
And write a much simpler program:
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("233168\n");
return 0;
}
You could change your ifs:
if ((start % 3 == 0) || (start % 5 == 0))
sum += start;
start ++;
and don´t forget to initialize your sum with zero and start with one.
Also, change the while condition to < 1000.
You would be much better served by a for loop, and combining your conditionals.
Not tested:
int main()
{
int x;
int sum = 0;
for (x = 1; x <= 1000; x++)
if (x % 3 == 0 || x % 5 == 0)
sum += x;
printf("%d\n", sum);
return 0;
}
The answers are all good, but won't help you learn C.
What you really need to understand is how to find your own errors. A debugger could help you, and the most powerful debugger in C is called "printf". You want to know what your program is doing, and your program is not a "black box".
Your program already prints the sum, it's probably wrong, and you want to know why. For example:
printf("sum:%d start:%d\n", sum, start);
instead of
printf("%d\n", sum);
and save it into a text file, then try to understand what's going wrong.
does the count start with 1 and end with 999?
does it really go from 1 to 999 without skipping numbers?
does it work on a smaller range?
Eh right, well i can see roughly where you are going, I'm thinking the only thing wrong with it has been previously mentioned. I did this problem before on there, obviously you need to step through every multiple of 3 and 5 and sum them. I did it this way and it does work:
int accumulator = 0;
int i;
for (i = 0; i < 1000; i += 3)
accumulator += i;
for (i = 0; i < 1000; i +=5) {
if (!(i%3==0)) {
accumulator += i;
}
}
printf("%d", accumulator);
EDIT: Also note its not 0 to 1000 inclusive, < 1000 stops at 999 since it is the last number below 1000, you have countered that by < 1001 which means you go all the way to 1000 which is a multiple of 5 meaning your answer will be 1000 higher than it should be.
You haven't said what the program is supposed to do, or what your problem is. That makes it hard to offer help.
At a guess, you really ought to initialize start and sum to zero, and perhaps the printf should be outside the loop.
Really you need a debugger, and to single-step through the code so that you can see what it's actually doing. Your basic problem is that the flow of control isn't going where you think it is, and rather than provide correct code as others have done, I'll try to explain what your code does. Here's what happens, step-by-step (I've numbered the lines):
1: while (start < 1001) {
2: if (start % 3 == 0) {
3: sum = sum + start;
4: start += 1;
5: }
6: else {
7: start += 1;
8: }
9:
10: if (start % 5 == 0) {
11: sum = sum + start;
12: start += 1;
13: }
14: else {
15: start += 1;
16: }
17: printf("%d\n", sum);
18: }
line 1. sum is 0, start is 0. Loop condition true.
line 2. sum is 0, start is 0. If condition true.
line 3. sum is 0, start is 0. sum <- 0.
line 4. sum is 0, start is 0. start <- 1.
line 5. sum is 0, start is 1. jump over "else" clause
line 10. sum is 0, start is 1. If condition false, jump into "else" clause.
line 15. sum is 0, start is 1. start <- 2.
line 16 (skipped)
line 17. sum is 0, start is 2. Print "0\n".
line 18. sum is 0, start is 2. Jump to the top of the loop.
line 1. sum is 0, start is 2. Loop condition true.
line 2. sum is 0, start is 2. If condtion false, jump into "else" clause.
line 7. sum is 0, start is 2. start <- 3.
line 10. sum is 0, start is 3. If condition false, jump into "else" clause.
line 15. sum is 0, start is 3. start <- 4.
line 17. sum is 0, start is 4. Print "0\n".
You see how this is going? You seem to think that at line 4, after doing sum += 1, control goes back to the top of the loop. It doesn't, it goes to the next thing after the "if/else" construct.
You have forgotten to initialize your variables,
The problem with your code is that your incrementing the 'start' variable twice. This is due to having two if..else statements. What you need is an if..else if..else statement as so:
if (start % 3 == 0) {
sum = sum + start;
start += 1;
}
else if (start % 5 == 0) {
sum = sum + start;
start += 1;
}
else {
start += 1;
}
Or you could be more concise and write it as follows:
if(start % 3 == 0)
sum += start;
else if(start % 5 == 0)
sum += start;
start++;
Either of those two ways should work for you.
Good luck!
Here's a general solution which works with an arbitrary number of factors:
#include <stdio.h>
#define sum_multiples(BOUND, ...) \
_sum_multiples(BOUND, (unsigned []){ __VA_ARGS__, 0 })
static inline unsigned sum_single(unsigned bound, unsigned base)
{
unsigned n = bound / base;
return base * (n * (n + 1)) / 2;
}
unsigned _sum_multiples(unsigned bound, unsigned bases[])
{
unsigned sum = 0;
for(unsigned i = 0; bases[i]; ++i)
{
sum += sum_single(bound, bases[i]);
for(unsigned j = i + 1; bases[j]; ++j)
sum -= sum_single(bound, bases[i] * bases[j]);
}
return sum;
}
int main(void)
{
printf("%u\n", sum_multiples(999, 3, 5));
return 0;
}