# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
85385 | DiegoGarcia | Nadan (COCI18_nadan) | C++14 | 2 ms | 620 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <stdio.h>
#define ll long long
ll n,k, ans[1000],each, res;
int main()
{
//los n_i son distintos enteros positivos que suman k y siempre se puede => k >= n*(n+1)/2
scanf("%lld %lld",&k,&n);
k -= ((n*(n+1))/2);
each = k/n;
res = k - each*n;
for( ll i=n-1; i >= 0; i-- )
{
ans[i] = each + i + 1;
if( res-- > 0 ) ans[i]++;
}
for(ll i=0; i<n; i++ ) printf("%lld ",ans[i]);
return 0;
}
Compilation message (stderr)
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