• Codeforces Round #672 (Div. 2)


    题目传送门

    A. Cubes Sorting

    题意: 询问逆序对个数是否不大于$ frac{n * (n - 1)}{2} - 1 $

    最不理想的情况就是单调下降序列,此时的逆序对个数为$ frac{n * (n - 1)}{2} $,只要不为单调下降序列就满足题意

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define rep(i, a, b) for (register int i = a; i <= b; i++)
     
    int n, a[50010];
    int b[50010];
    inline void solve(int T) {
        cin >> n; 
        rep(i, 1, n) cin >> a[i];
        rep(i, 2, n) {
            if(a[i] >= a[i - 1]) {
                cout << "YES
    ";
                return;
            }
        }
        cout << "NO
    ";
        
    }
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     
        // freopen("in.txt", "r", stdin);
        // freopen("out2.txt", "w", stdout);
     
        int T = 1;
        cin >> T;
        rep(i, 1, T) solve(i);
    }
    View Code

     

    B. Rock and Lever

    题意: 已知序列 ${a_i}$, 求$a_i & a_j geq a_i oplus a_j, i < j$的个数

    容易发现,当两个数转化成二进制位数相同的时候满足条件,简单的计算一下就可

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define rep(i, a, b) for (register int i = a; i <= b; i++)
     
    ll n;
    ll a[100010];
    ll cnt[50];
    ll ans;
    inline void solve(int T) {
     
        cin >> n;
        memset(cnt, 0, sizeof(cnt));
        rep(i, 1, n) {
            cin >> a[i];
            int tmp = 0;
            while(a[i]) {
                tmp++;
                a[i] /= 2;
            }
            cnt[tmp]++;
        }
        ans = 0;
        rep(i, 1, 49) ans += cnt[i] * (cnt[i] - 1) / 2;
        cout << ans << endl;
     
    }
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     
        // freopen("in.txt", "r", stdin);
        // freopen("out2.txt", "w", stdout);
     
        int T = 1;
        cin >> T;
        rep(i, 1, T) solve(i);
    }
    View Code

    C1. Pokémon Army (easy version)

    题意: 从序列中选择若干个数,最大化$a_{b_1}-a_{b_2}+a_{b_3}-a_{b_4}...$

    考虑dp,$dp[i[[0]$表示加上第$i$个数的最大值,$dp[i][1]$ 表示减去第$i$个数的最大值

    所以$dp[i][0]=max(dp[j][1]) + a[i], dp[i][1]=max(dp[j][0])-a[i],j<i$

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define rep(i, a, b) for (register int i = a; i <= b; i++)
     
    ll n, q;
    ll a[300010];
    ll dp[300010][2], mx[2], ans;
    inline void solve(int T) {
        cin >> n >> q;
        rep(i, 1, n) cin >> a[i];
        mx[0] = mx[1] = ans = 0;
        rep(i, 1, n) {
            dp[i][0] = mx[1] + a[i];
            dp[i][1] = mx[0] - a[i];
            mx[0] = max(mx[0], dp[i][0]);
            mx[1] = max(mx[1], dp[i][1]);
            ans = max(ans, mx[0]);
        }
        cout << ans << endl;
    }
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     
        // freopen("in.txt", "r", stdin);
        // freopen("out2.txt", "w", stdout);
     
        int T = 1;
        cin >> T;
        rep(i, 1, T) solve(i);
    }
    View Code

     

    C2. Pokémon Army (hard version)

    有个贪心的结论,答案为序列差分数组正数的和。

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define rep(i, a, b) for (register int i = a; i <= b; i++)
     
    ll n, q;
    ll a[300010];
    ll ans;
    inline void solve(int T) {
        cin >> n >> q;
        ans = a[n + 1] = 0;
        rep(i, 1, n) {
            cin >> a[i];
            if(a[i] - a[i - 1] > 0) ans += a[i] - a[i - 1];
        }
        cout << ans << endl;
        ll l, r;
        rep(i, 1, q) {
            cin >> l >> r;
            ll tmp1 = a[l], tmp2 = a[r];
            ans -= max(a[l + 1] - tmp1, 0ll) + max(tmp1 - a[l - 1], 0ll);
            a[l] = tmp2;
            ans += max(a[l + 1] - tmp2, 0ll) + max(tmp2 - a[l - 1], 0ll);
            ans -= max(a[r + 1] - tmp2, 0ll) + max(tmp2 - a[r - 1], 0ll);
            a[r] = tmp1;
            ans += max(a[r + 1] - tmp1, 0ll) + max(tmp1 - a[r - 1], 0ll);
            cout << ans << endl;
        }
    }
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
     
        // freopen("in.txt", "r", stdin);
        // freopen("out.txt", "w", stdout);
     
        int T = 1;
        cin >> T;
        rep(i, 1, T) solve(i);
    }
    View Code

    D. Rescue Nibel!

    题意: 选择$k$个等使得这$k$个灯至少在某一时刻同时亮着

    枚举时刻,在当前时刻才亮的灯与前一个时刻亮到当前时刻的灯进行组合

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define rep(i, a, b) for (register int i = a; i <= b; i++)
     
    struct bulb {
        int l, r;
        bool operator<(bulb & f) const {
            if(l == f.l) return r < f.r;
            return l < f.l;
        }
    }a[300010];
    ll n, k, cnt, ans;
    priority_queue<int, vector<int>, greater<int> > q;
     
    ll mod = 998244353;
    ll ksm(ll x, ll b)
    {
        ll res = 1;
        for (; b; b >>= 1, x = x * x % mod)
            if (b & 1)
                res = res * x % mod;
        return res;
    }
    ll fac[300010], inv[300010];
    ll C(ll f, ll j) { return fac[f] * inv[j] % mod * inv[f - j] % mod; }
     
    inline void solve(int T) {
     
        cin >> n >> k;
        rep(i, 1, n) cin >> a[i].l >> a[i].r;
        sort(a + 1, a + n + 1);
        for(int i = 1; i <= n; ) {
            int flag = a[i].l, tmp = 0;
            while(flag == a[i].l) {
                q.push(a[i].r);
                tmp++;
                i++;
            }
            while(q.top() < flag) {
                q.pop();
                cnt--;
            }
            rep(j, max(1ll, k - cnt), k) if(tmp >= j) {
                ans += C(cnt, k - j) * C(tmp, j) % mod;
                ans %= mod;
            }
            else break;
            cnt += tmp;
        }
        cout << ans << endl;
     
    }
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        
        fac[0] = inv[0] = 1;
        rep(i, 1, 300000) fac[i] = fac[i - 1] * i % mod;
        inv[300000] = ksm(fac[300000], mod - 2);
        for (int i = 299999; i > 0; i--)
            inv[i] = inv[i + 1] * (i + 1) % mod;
     
        // freopen("in.txt", "r", stdin);
        // freopen("out2.txt", "w", stdout);
     
        int T = 1;
        // cin >> T;
        rep(i, 1, T) solve(i);
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/likunhong/p/13728716.html
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