I managed to solve this, below is my solution :
public class ProblemA001k {
public static void main(String[] args) {
System.out.println("Sum from 1" + " to " + divQ + ":" + sum2);
System.out.println();
divQ += q;
newQ += q;
sum1 = 0;
sum2 = 0;
}
key.close();
}
}
Now I was told to modify my solution so that it uses ONLY ONE LOOP. I have 3 loops in the code above, even when I tried using only 2 loops I struggled. but ONE LOOP ? I don't know how to improve my code. Please help me.
This is a Mathematic
problem.
If you know that you can find the sum of all integers from 1
to X
, you just need to do X * (X+1) / 2
.
You can find all the batch values easily.
Sum from 1 to 400: 80200
Sum from 401 to 450: 21275
Sum from 1 to 450: 101475
Will be found like this :
450 * 451 / 2 = 101475 (1 to 450)
400 * 401 / 2 = 80200 (1 to 400)
101475 - 80200 = 21275 (401 to 450)
With this, you can limit the loop to just calculate the values from q
to n
by incrementing by q
And a quick code to do it :
static void sum(int n, int q){
int i = q;
int sum, tmp=0;
while(i < n){
sum = i * (i+1) / 2;
System.out.println(String.format("Sum from %d to %d : %d", i-q+1 , i, sum - tmp));
System.out.println(String.format("Sum from %d to %d : %d", 1, i, sum));
tmp = sum;
i += q;
}
}
And I run it with
public static void main(String[] args){
sum(500, 50);
}
to have this result
Sum from 1 to 50 : 1275
Sum from 1 to 50 : 1275
Sum from 51 to 100 : 3775
Sum from 1 to 100 : 5050
Sum from 101 to 150 : 6275
Sum from 1 to 150 : 11325
Sum from 151 to 200 : 8775
Sum from 1 to 200 : 20100
Sum from 201 to 250 : 11275
Sum from 1 to 250 : 31375
Sum from 251 to 300 : 13775
Sum from 1 to 300 : 45150
Sum from 301 to 350 : 16275
Sum from 1 to 350 : 61425
Sum from 351 to 400 : 18775
Sum from 1 to 400 : 80200
Sum from 401 to 450 : 21275
Sum from 1 to 450 : 101475
The good think with this solution is the number of loop, this will increment by q
instead of 1
Note : The solution is a quick implementation, this could be done better.
EDIT :
Thanks to Margaret Bloom in the comments to point out the name of this formula :) For more information, you are welcome to look at Triangular Number
This should do it:
int totalSum = 0;
int batchSum = 0;
for (int i = 1; i <= n; i++) {
totalSum += i;
batchSum += i;
if (i % q == 0) {
System.out.println("Sum from " + (i - q + 1) + " to " + i + ":" + batchSum);
System.out.println("Sum from 1 to " + i + ":" + totalSum);
batchSum = 0;
}
}
Edit: The better Math way:
int lastTotalSum = 0;
for (int i = 1; i <= n / q; i++ ) {
int top = i * q;
int totalSum = top * (top + 1) / 2;
int batchSum = totalSum - lastTotalSum;
System.out.println("Sum from " + (top - q + 1) + " to " + top + ":" + batchSum);
System.out.println("Sum from 1 to " + top + ":" + totalSum);
lastTotalSum = totalSum;
}
I found a nice solution with java8 Streams:
int n=1000;
int q=50;
int length = n/q -1;
int[] previousSum={0};
IntStream.range(0, length).map(i -> (i+1)*q).forEach(s -> {
int sum=(s*(s+1))/2;
int batch = sum - previousSum[0];
previousSum[0] = sum;
System.out.println("Sum from " + (s - q + 1) + " to " + s + ":" + batch);
System.out.println("Sum from 1 to " + s + ":" + sum);
});
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