A.復制雞

思路：比較簡單，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.好伙計猜拳

思路：這道題和最長子序列相似，在符合時加1，而這里是加懲罰值，并且又有交換的條件，所以開二維數組來進行動態轉移。

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.數列之和

思路：這道題可以打表觀察出來規律，之前以為會是加2和加4的規律，發現并沒有什么規律，再仔細觀察，查看缺失值是哪些，發現是,所以就簡單了，二分一下就能做出來，并注意二分的左右取值，1e18大概是,以為k的范圍到1e18,其中并不可能缺失一半，所以假設k的范圍為2e18，所以來確定取值，如果取值太大后對2取指數會超范圍。

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.薪得體會

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.小雞的排列構造

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.小雞的排列構造的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.鐵刀磨成針

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.雞翻題

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.變雞器

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;
}

思路待更新。。。