What does k*1LL means? K is any integer

Asked Nov 08 '17 at 13:50

Active Nov 08 '17 at 13:53

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I was doing a problem it was related to normal DFS, few of test cases were not passing so i tried different things:

1: (long long)(d * (n - sz[u])); // not passing

2: d* 1LL* (n-sz[u]); //passing all

const int N = 100000;
vector <int> adj[N + 1];
bool visited[N + 1];
int sz[N + 1];
int n;
long long num, den;

void dfs(int u, int d) {
visited[u] = 1;
sz[u] = 1;
for(auto v: adj[u]) {
   if(!visited[v]) {
      dfs(v, d + 1);
      sz[u] += sz[v];
   }
}
if(u != 1) {
   num += (long long)(d  *  (n - sz[u])); //prob lies here
   den += d;
}

}

I think both are essentially same, but clearly they are not. can someone help ?

edit: what exactly k = d*1LL means ?? what would be the value of k ?

edited Nov 08 '17 at 13:53

asked Nov 08 '17 at 13:50

ritesh chauhan

what is the difference between the 2 statements ?? – ritesh chauhan Nov 08 '17 at 13:56
`(long long)(d * (n - sz[u]))` only casts the **result** to `long long`. It doesn't turn the operands into `long long` integers, and doesn't do the computation with `long long` integers. – StoryTeller - Unslander Monica Nov 08 '17 at 13:59
In 1) the calculations are performed using `int`, possibly overflowing. Then the result is converted to `long long`. In 2) the precence of `1LL` causes the other values to also be converted to the common type `long long` *before* doing the calculations. – Bo Persson Nov 08 '17 at 14:00
got the point !! thankyou all – ritesh chauhan Nov 08 '17 at 14:02
If you want your questions to be better received take the [tour], and read [how to ask](https://stackoverflow.com/help/asking). – StoryTeller - Unslander Monica Nov 08 '17 at 14:06

What does k*1LL means? K is any integer

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