This blog is intended to discuss many things of which algorithmic problems are a part.
So lets begin with a fairly simple problem Create Collections on spoj:
The problem can easily be solved by hashing as the constraints are really poor. By hashing, we can find out for each number whether a corresponding double number exists or not. Rest is simple :) Below is the accepted code for the problem:
So lets begin with a fairly simple problem Create Collections on spoj:
The problem can easily be solved by hashing as the constraints are really poor. By hashing, we can find out for each number whether a corresponding double number exists or not. Rest is simple :) Below is the accepted code for the problem:
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| #include<iostream> | |
| #include<cstdio> | |
| #include<algorithm> | |
| #include<cstdlib> | |
| #include<cstring> | |
| using namespace std; | |
| int hash[10000000]; | |
| int main() | |
| { | |
| int t, n, i, a[103], count; | |
| scanf("%d", &t); | |
| while(t--) { | |
| count=0; | |
| memset(hash, 0, sizeof(hash)); | |
| scanf("%d", &n); | |
| for(i=0; i<n; i++) { | |
| scanf("%d", &a[i]); | |
| hash[a[i]]+=1; | |
| } | |
| sort(a, a+n); | |
| for(i=0; i<n; i++) { | |
| if(hash[a[i]]>0 && hash[2*a[i]]>0) { | |
| hash[a[i]]--; | |
| hash[2*a[i]]--; | |
| count++; | |
| } | |
| } | |
| printf("%d\n", count); | |
| } | |
| return 0; | |
| } |
