AtCoder Beginner Contest 429

A - Too Many Requests

AC 代码

    int n, m;

    void solve(){
        cin >> n >> m;
        for(int i = 1; i <= n; i++){
            cout << (i <= m ? "OK" : "Too Many Requests") << endl;
        }
        return ;
    }

B - N-1

AC 代码

    int sum = 0;
    set<int> st;
    int n, m;
    int a[N];

    void solve(){
        cin >> n >> m;
        for(int i = 1;i <= n; i ++){
            cin >> a[i];
            sum += a[i];
            st.insert(a[i]);
        }
        for(int i = 0; i <= n; i ++){
            if(st.find(sum - m) != st.end()){
                cout << "Yes" << endl;
                return ;
            }
        }
        cout << "No" << endl;
        return ;
    }

C - Odd One Subsequence

AC 代码

    int n;
    int a[N];
    map<int, int> mp;
    int ans = 0;

    void solve(){
        cin >> n;
        for(int i = 1; i <= n; i ++){
            cin >> a[i];
            mp[a[i]] ++;
        }
        for(auto i : mp){
            if(i.second >= 2){
                ans += (i.second - 1) * i.second / 2 * (n - i.second);
            }
        }
        cout << ans << endl;
        return ;
    }

D - On AtCoder Conference

AC 代码

    int n, m, c;
    int a[N];
    int ans = 0;

    void solve(){
        cin >> n >> m >> c;
        for(int i = 1;i <= n; i ++){
            cin >> a[i];
        }
        sort(a + 1, a + n + 1);
        for(int i = 1;i <= n; i ++){
            a[n + i] = a[i] + m;
        }
        
        for(int i = 1; i <= n; i++){
            int p = upper_bound(a + 1, a + 1 + 2*n, a[i + c])-a-1;
    		ans += (p - i) * (a[i+1] - a[i]);
        }
        cout << ans << endl;
        return ;
    }