| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 375211 | kiomi | Sifra (COCI21_sifra) | C++17 | 1 ms | 364 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.
//order_of_key(k): Number of items strictly smaller than k .
//find_by_order(k): K-th element in a set (counting from zero).
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define full(x,n) x,x+n+1
#define full(x) x.begin(),x.end()
#define f first
#define s second
#define len(x) (int)x.size()
//logx(a^n)=loga(a^n)/logx(a)
//logx(a*b)=logx(a)+logx(b)
//logx(y)=log(y)/log(x)
//logb(n)=loga(n)/loga(b)
#define ordered_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
//(a+b)^n=sum of C(n,i)*a^i*b^(n-i) 0<=i<=n
//(a-b)^n=sum of C(n,i)*a^i*b^(n-i) for even i-for odd i
// 1/b%mod=b^(m-2)%mod
// (a>>x)&1==0
// a^b=(a+b)-2(a&b)
typedef double db;
typedef long long ll;
//sum of squares n*(n+1)*(2n+1)/6
//sum of cubes [n*(n+1)/2]^2
//sum of squares for odds n*(4*n*n-1)/3
//sum of cubes for odds n*n*(2*n*n-1)
const int ary=1e6+5;
const int mod=1e9+7;
const ll inf=1e18;
using namespace std;
using namespace __gnu_pbds;
string s;
int cnt[ary],ans;
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>s;
	int knt=0;
	for(int i=0;i<len(s);i++){
		if(s[i]>='0'&&s[i]<='9'){
			knt=knt*10+(s[i]-'0');
			if(i==len(s)-1||(s[i+1]<'0'||s[i+1]>'9')){
				cnt[knt]++;
				if(cnt[knt]==1){
					ans++;
				}
				knt=0;
			}
		}
	}
	cout<<ans;
}
Compilation message (stderr)
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