Comet OJ - Contest # 15 problem solution

A double eleven preferential (simple version)

 n  <=  1e19,   1e9 > 1(8)

https://www.cometoj.com/contest/79/problem/A?problem_id=4198

#include<bits/stdc++.h>

using namespace std;

int main(){
    int t; cin >> t; 
    while(t--) {
        int cnt1 = 0; 
        int n;cin >> n;
        int pr[] = {1,11,111,1111,11111,111111,1111111,11111111,111111111};
        for(int i = 8; i >= 0; i--) {
            Chntl + = N / Q [i];
            Neither % = Q [i];
        }
        if(cnt1 > 9) cout << "Impossible\n";
        else cout << cnt1 << endl;
    }
    return 0;
}

D version difficulty

https://www.cometoj.com/contest/79/analysis  // dm get together online Topic, passing through do not miss

 May be assumed that a particular band, 1111111111 ............

 This is then converted to decimal number hexadecimal, for example 234--> 2,1,1 (2 * 111 + 1 * 11 + 1) 

 Hexadecimal digits in the sum of the number of "good number" of need, we will find that when other digits except the last one is greater than 9:00, can be converted to the smaller band
 for example 11,0 0 -> 1,0,10,0 10,0,1,0 -> 1,0,0,0,10  
 then we can as i i th binary, hexadecimal take contribution under this

 11,0,0 -> 1,0,10,0 (3 * 11 * 4 = 331 = 14 + 10 * 1) 10,0,1,0 -> 1,0,0,0,10 (10 * 4 + 2 = 421 * 5 + 10 = 15) 
 will find that when there are more than 9 digits (not the last one) into a smaller band contribution is smaller, so we can have him into smaller therefore, in line with selection greedy nature
 Conclusion: when a decimal number is represented by "good number", with the greatest possible "good number" of the current meet said minimum number "good number" spent

 

 

B when we are together, concentric

https://www.cometoj.com/contest/79/problem/B?problem_id=4199

#include <bits / STDC ++. H>
 #define LL Long Long
 // violence enumerate each coordinate is used as the center can have several points of equal distance from the center, ANS = F + [CNT1]
 // Each another point as the center point of the array as the distance to the center and then seek a few sort, so the floating point accuracy errors will change the function from the square root can not ll ac 
distance // may be obtained by the map memory per , 

a using  namespace std;
ll dist(ll x1, ll y1, ll x2, ll y2){
    return ((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main(){
    int t; 
    t = 1;
    //cin >> t; 
    while(t--) {
        int n;cin >> n;
        int ans = 0;
        //vector<pair<int, int> > dot(n+1);
        ll dx[2020], dy[2020];
        for(int i = 0; i < n; i++) cin >> dx[i] >> dy[i];
        for(int i = 0; i < n; i++) {
            int x = i;
            vector<ll> dis;
            for(int j = 0; j < n; j++){
                if(i == j) continue;
                ll distan = dist(dx[i], dy[i], dx[j], dy[j]);
                dis.push_back(distan);
        //cout << "i = " << i << "  dis = " << distan << endl << endl; 
            }
            sort(dis.begin(), dis.end());
            ll y = -1; int cnt1 = 1;
            for(int j = 0; j < n-1; j++){
    //cout << dis[j] << "  y = " << y << "  cnt1 = " << cnt1 << endl << endl;
                if(dis[j] == y) cnt1++;
                else {
                    years + = (CNT1 * (cnt1- 1 ) * (cnt1- 2 ) / 6 );
                    y = tell [j]; CNT1 = 1 ;
                }
            }
    //cout << "ans = " << ans << " cnt1=  "<< cnt1<< " f[cnt1]" << f[cnt1] << endl << endl;
            ans += (cnt1*(cnt1-1)*(cnt1-2)/6);
    //cout << "!111\n";
        }
        cout << ans << endl;
    }
    return 0;
}

 

#include <bits / STDC ++ H.>
 #define LL Long Long
 // read the explanations, a case slightly to optimize 
the using  namespace STD;
ll dist(ll x1, ll y1, ll x2, ll y2){return ((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}
int main(){
    int t; 
    t = 1;
    while(t--) {
        int n;cin >> n;
        int ans = 0;
        //vector<pair<int, int> > dot(n+1);
        ll dx[2020], dy[2020];
        for(int i = 0; i < n; i++) cin >> dx[i] >> dy[i];
        for(int i = 0; i < n; i++) {
            vector<ll> dis;
            for(int j = 0; j < n; j++){
                dis.push_back(dist(dx[i], dy[i], dx[j], dy[j]));
            }
            sort(dis.begin(), dis.end());
            for(int j = 0, k; j < n; j = k){
                for(k = j; dis[k] == dis[j]; k++);
                years + = (kj) * (kj- 1 ) * (kj- 2 ) / 6 ;
            }
        }
        cout << ans << endl;
    }
    return 0;
}