2025牛客寒假算法基础集训营6

A.复制鸡

思路：比较简单，略。

void solve()
{
    int n, m, k;
    cin >> n;
    int last = -1, ans = 0;
    for (int i = 0; i<n; i++){
        int x;
        cin >> x;
        if (x != last){
            ans++;
        }
        last = x;
    }
    cout << ans << endl;
}

B.好伙计猜拳

思路：这道题和最长子序列相似，在符合时加1，而这里是加惩罚值，并且又有交换的条件，所以开二维数组来进行动态转移。

void solve()
{
    int n, c1, c2;
    cin >> n >> c1 >> c2;
    vector<int> a(n+1), b(n+1);
    for (int i = 1; i<=n; i++){
        cin >> a[i] >> b[i];
    }
    vector<vector<int>> dp(n+1, vector<int>(2, INF));
    dp[0][0] = 0;
    int ans = INF;
    for (int i = 1; i<=n; i++){
        for (int j = 0; j<i; j++){
            if (a[j] <= a[i] && b[j] <= b[i]) // (aj, bj) -> (ai, bi)
                dp[i][0] = min(dp[i][0], dp[j][0]+c1*(i-j-1));
            if (a[j] <= b[i] && b[j] <= a[i]) // (bj, aj) -> (ai, bi)
                dp[i][0] = min(dp[i][0], dp[j][1]+c1*(i-j-1));
        }
        for (int j = 0; j<i; j++){    
            if (a[j] <= b[i] && b[j] <= a[i]) // (aj, bj) -> (bi, ai)
                dp[i][1] = min(dp[i][1], dp[j][0]+c1*(i-j-1));
            if (b[j] <= b[i] && a[j] <= a[i]) // (bj, aj) -> (bi, ai)
                dp[i][1] = min(dp[i][1], dp[j][1]+c1*(i-j-1));
        }
        dp[i][1] += c2;
        ans = min({ans, dp[i][0]+c1*(n-i), dp[i][1]+c1*(n-i)});
    }
    cout << ans << endl;
}

C.数列之和

思路：这道题可以打表观察出来规律，之前以为会是加2和加4的规律，发现并没有什么规律，再仔细观察，查看缺失值是哪些，发现是,所以就简单了，二分一下就能做出来，并注意二分的左右取值，1e18大概是,以为k的范围到1e18,其中并不可能缺失一半，所以假设k的范围为2e18，所以来确定取值，如果取值太大后对2取指数会超范围。

int qpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res *= a) ; 
        a *= a ;
        b >>= 1;
    }
    return res;
}
 
void solve()
{
    int n, m, k;
    cin >> n;
    if (n == 1){
        cout << 2 << endl;
        return ;
    }
    auto check = [&](int x)->bool
    {
        int p = qpow(2, x);
        return p/2-(x-1) >= n;
    };
    int l = 0, r = 125;
    while (l+1 < r){
        int mid = (l+r) >> 1;
        if (check(mid)) r = mid;
        else l = mid;
    }
    cout << 2*(r+n-2) << endl;
}

F.薪得体会

void solve()
{
	int n, m, k;
	cin >> n;
	vector<node> s(n+1);
	for (int i = 1; i<=n; i++){
		cin >> s[i].a >> s[i].b;
		s[i].sum = s[i].a + s[i].b;
	}
	int ans = 0;
	sort(s.begin()+1, s.end(), [&](node x, node y){
		return x.sum < y.sum;
	});
	vector<int> mx(n+1);
	for (int i = 1; i<=n; i++) mx[i] = max(mx[i-1], s[i].a+s[i].b);
	for (int i = 1; i<=n; i++){
		if (ans >= s[i].a) ans = max(ans, s[i].sum);
		else ans = max(ans, s[i-1].sum);
		ans = max(ans, s[i].a);
	}
	cout << ans << endl;
}

H.小鸡的排列构造

void solve()
{
    int n, m;
    cin >> n >> m;
    int l, r, c;
    for (int i = 0; i<m; i++){
        cin >> l >> r >> c;
    }
    if ((l-r+1) % 2 == 0){
        for (int i = n; i>=1; i--){
            cout << i << " \n"[i==1];
        }
        return ;
    }
    else{
        vector<int> a(n+1);
        iota(a.begin()+1, a.end(), 1);
        sort(a.begin()+1, a.end(), greater<int>());
        for (int i=n; i>=2; i-=2){
            swap(a[i], a[i-1]);
        }
        for (int i = 1; i<=n; i++){
            cout << a[i] << " \n"[i==n];
        }
    }
}

I.小鸡的排列构造的checker

struct query{
    int l, r, c;
};
class Fenwick{
public:
    vector <int> tree;
    int n;
    Fenwick (){}
    Fenwick (int x){
        init(x);
    }
    void init(int x){
        n = x;
        tree.resize(n + 1);
    }
    int lowbit(int x){
        return x & (-x);
    }
    void add(int x, int k){
        for(; x <= n; x += lowbit(x))
            tree[x] += k;
    }
    int qsum(int x){
        int ans = 0;
        for(; x; x -= lowbit(x))
            ans += tree[x];
        return ans;
    }
    void clean(){
        for(int i = 1; i <= n; i ++) tree[i] = 0;
    }
};
void solve()
{
    int n, m, k;
    cin >> n >> m;
    Fenwick fen(n);
    vector<int> pos(n+1), val(n+1);
    for (int i = 1; i <= n; i++) {
        cin >> val[i];
        pos[val[i]] = i;
    }
    vector<query> que(m+1);
    vector<vector<int>> q(n+1);
    for (int i = 0; i<m; i++){
        cin >> que[i].l >> que[i].r >> que[i].c;
        q[val[que[i].c]].push_back(i);
    }
    vector<int> ans(m);
    for (int i = 1; i<=n; i++){
        for (int j: q[i]) ans[j] = fen.qsum(que[j].r)-fen.qsum(que[j].l-1)+que[j].l;
        fen.add(pos[i], 1);
    }
    for (int i=0; i<m; i++){
        cout << ans[i] << "\n";
    }
}

J.铁刀磨成针

void solve()
{
    int n, x, y;
    cin >>n >> x >> y;
    auto dc = [&](int s, int xs)->int
    {
        int res = (s+(s-xs+1))*xs/2;
        return res;
    };
    int ans = 0;
    for (int i = 1; i<=min(n, y); i++){
        int now = x+i;
        int sum = now*(min(n, y)-i);
        sum += dc(now, min(now, n-min(n, y)+1));
        ans = max(ans, sum);
    }
    cout << ans << endl;
}

K.鸡翻题

void solve()
{
    int n, x, y;
    cin >> x >> y;
    if (y % 2 == 0){
        cout << "NO" << endl;
        return ;
    }
    if (x % 2 == 0){
        if (y % 4 == 1){
            cout << "YES" << endl;
            return ;
        }
        else {
            cout << "NO" << endl;
            return ;
        }
    }
    else {
        if (y % 4 == 3){
            cout << "YES" << endl;
            return ;
        }
        else {
            cout << "NO" << endl;
            return ;
        }
    }
}

 L.变鸡器

void solve()
{
    int n, m, k;
    cin >> n;
    string s, st = "CHICKEN";
    cin >> s;
    map<char, int> mp;
    mp['C'] = -2, mp['H'] = -1, mp['I'] = -1, mp['K']=-1, mp['E']=-1, mp['N'] = -1;
    int num = 0;
    for (int i = 0; i<n; i++){
        if (s[i] == st[num]) num++;
        mp[s[i]]++;
    }
    if (num != 7){
        cout << "NO" << endl;
        return ;
    }
    if ((n-num)%2){
        cout << "NO" << endl;
        return ;
    }
    vector<int> sb;
    for (auto [x, nn] : mp){
        sb.push_back(nn);
    }
    int max_num = *max_element(sb.begin(), sb.end());
    int sum = accumulate(sb.begin(), sb.end(), 0ll);
    if (max_num <= sum/2){
        cout << "YES" << endl;
    }
    else cout << "NO" << endl;
}