I was recently wondering what complexity loop can be used in less than 1 sec. For the same, I took one of my previously solved problems and added an extra loop but, even after I increase the looping to as high as 10^15, I get Accepted within 15ms. How could this happen?

Attached is a screenshot of my solution which includes a 10^15 loop which gets accepted within 15ms.

Problem link: http://codeforces.me/problemset/problem/371/C

Solution:

include <bits/stdc++.h>

using namespace std;

typedef pair<int,int> II; typedef vector< II > VII; typedef vector VI; typedef vector< VI > VVI; typedef long long int LL;

define PB push_back

define MP make_pair

define F first

define S second

define SZ(a) (int)(a.size())

define ALL(a) a.begin(),a.end()

define SET(a,b) memset(a,b,sizeof(a))

define din(n) int n; scanf("%d",&n)

define dout(n) printf("%d\n",n)

define llin(n) long long n; scanf("%lld",&n)

define llout(n) printf("%lld\n",n)

define strin(s,l) char s[l]; scanf("%s",s)

define strout(n) printf("%s\n",n)

define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)

define nl '\n'

define TRACE

ifdef TRACE

define trace(...) f(#VA_ARGS__, VA_ARGS)

template void __f(const char* name, Arg1&& arg1){ cerr << name << " : " << arg1 << std::endl; } template <typename Arg1, typename... Args> void f(const char* names, Arg1&& arg1, Args&&... args){ const char* comma = strchr(names + 1, ',');cerr.write(names, comma — names) << " : " << arg1<<" | ";f(comma+1, args...); }

else

define trace(...)

endif

bool cmpManh(const std::pair<long long,long long>& l, const std::pair<long long,long long>& r) { return ((llabs(l.F) + llabs(l.S)) < (llabs(r.F) + llabs(r.S))); }

int main(void) { string str; cin>>str; LL reqb = count(ALL(str),'B'); LL reqs = count(ALL(str),'S'); LL reqc = count(ALL(str),'C'); LL b,s,c; LL pb,ps,pc; cin>>b>>s>>c; cin>>pb>>ps>>pc; LL money; cin>>money; LL l = 0,h = 1e13,m; LL ab,ac,as,reqmon,reqmon2; LL abs,acs,ass,ans = 0; while(l<=h){ m = (l+h)/2; ab = (reqb*m — b)*pb; as = (reqs*m — s)*ps; ac = (reqc*m — c)*pc; abs = (reqb*(m+1) — b)*pb; ass = (reqs*(m+1) — s)*ps; acs = (reqc*(m+1) — c)*pc;

    reqmon = max(0LL,ab) + max(0LL,ac) + max(0LL,as);
    reqmon2 = max(0LL,abs) + max(0LL,acs) + max(0LL,ass);
    if(reqmon <= money && reqmon2>money){
       ans = m;
       break;
    }
    else if(reqmon <= money){
       l = m + 1;
    }
    else h = m - 1;

}
LL cc = 0;
for (LL i = 0; i < (long long)(1e15); ++i)
{
    // cc++;
    if(cc>0) cc++;
    else cc--;
    /* code */
}
cout<<ans<<nl;
return(0);

} PS: If this a bad place to ask for this query, please let me know where to post this. :)