Wilson Tan - 1 year ago 71

Java Question

I was attempting to solve this morning's Codeforces problem Div 2C: http://codeforces.com/contest/716/problem/C

This problem has the potential to loop up to 100,000 times so the parameter here can be up to 100,000. Loop seems to break when passing in 100,000 (and possibly earlier) and i is declared as an int:

`public void solve(int a) {`

double x = 2;

double y = 0;

double n = 0;

double target = 0;

double lcm = 0;

for (int i = 1; i <= a; i++) {

lcm = (i + 1) * i;

y = ((lcm * lcm) - x) / i;

n = (y * i) + x;

if (Math.sqrt(n) % (i + 1) == 0) {

x = Math.sqrt(n);

String answer = String.format("%.0f", y);

System.out.println("this is i: " + i);

System.out.println(answer);

}

}

}

Here is the relevant output:

`this is i: 46337`

99495281029892

this is i: 46338

99501722706961

this is i: 46340

99514606895203

this is i: 65535

32769

Doing a quick search on Stack overflow shows that the number 65535 is associated with a 16-bit unsigned int, but java uses 32bit ints. Changing the type to

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Answer Source

The following line generates an out of bounds `int`

before converting the result to `double`

:

```
lcm = (i + 1) * i;
```

The above is essentially the same as:

```
lcm = (double)((i + 1) * i);
```

or

```
int temp = (i + 1) * i;
lcm = (double) temp;
```

Instead try (first converting to double and then taking what is similar to a square):

```
lcm = (i + 1.0) * i;
```

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