B.
将$8\times8$的棋盘坐标 01 分解。定义棋盘的值为
$$ \oplus_{i, j}[a_{ij}]\cdot(8i+j) $$
其中$\oplus$为异或和，$a_{ij}$为$i$行$j$列是否为$1$。这样棋盘的每一格都唯一对应$[0, 63]$的一个数，且棋盘的值也是这样的一个数。
假设当前棋盘值为$v$，我们期望得到一个$[0,63]$的一个数$x$，我们只需将$x\mathrm{xor} v$对应的格子取反即可。
注意原题目要求的数域是$[1,64]$。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int N = 8; 6 7 bool a[N][N]; 8 9 int calc() { 10 int ret = 0; 11 for (int i = 0; i < N; ++i) 12 for (int j = 0; j < N; ++j) 13 if (a[i][j]) 14 ret ^= i * 8 + j; 15 return ret; 16 } 17 18 int main() { 19 ios::sync_with_stdio(false); 20 string name; 21 cin >> name; 22 if (name == "Mia") { 23 int t; 24 cin >> t; 25 while (t--) { 26 int num; 27 for (int i = 0; i < N; ++i) 28 for (int j = 0; j < N; ++j) 29 cin >> a[i][j]; 30 cin >> num; 31 num--; 32 int cur = calc(); 33 int tmp = num ^ cur; 34 a[tmp / 8][tmp % 8] ^= 1; 35 assert(calc() == num); 36 for (int i = 0; i < N; ++i) 37 for (int j = 0; j < N; ++j) 38 cout << a[i][j] << " \n"[j == N - 1]; 39 cout << "---\n"; 40 } 41 } else if (name == "Danila") { 42 int t; 43 cin >> t; 44 while (t--) { 45 int num; 46 for (int i = 0; i < N; ++i) 47 for (int j = 0; j < N; ++j) 48 cin >> a[i][j]; 49 string tmp; 50 cin >> tmp; 51 cout << calc() + 1 << endl; 52 } 53 } 54 }
E.
签到题。注意对$0$特殊处理。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 string tostr(unsigned x) { 6 string ret = ""; 7 while (x % 2 == 0) { 8 x >>= 1; 9 } 10 while (x) { 11 if (x & 1) { 12 ret += "1"; 13 } else { 14 ret += "0"; 15 } 16 x >>= 1; 17 } 18 return ret; 19 } 20 21 int main() { 22 // freopen("in", "r", stdin); 23 // freopen("out", "w", stdout); 24 ios::sync_with_stdio(false); 25 int T; 26 cin >> T; 27 while (T--) { 28 unsigned x, y; 29 cin >> x >> y; 30 if (y == 0) { 31 cout << "Yes\n"; 32 } else if (x == 0) { 33 cout << "No\n"; 34 } else { 35 string a = tostr(x), b = tostr(y); 36 if (a.find(b) == string::npos) { 37 cout << "No\n"; 38 } else { 39 cout << "Yes\n"; 40 } 41 } 42 } 43 }
F.
对所有边以边的贡献从大到小排序，依次贪心取边即可。这需要维护一个节点到根的最小值以及区间减法。（学傻了）所以用重链剖分维护。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef pair<int, int> pii; 6 7 const int MAXN = 100000 + 10; 8 9 struct Edge { 10 int v, n; 11 int64_t w; 12 Edge(int v=0, int n=0, int64_t w=0) : v(v), n(n), w(w) {} 13 } e[MAXN << 1]; 14 int G[MAXN], edgeCnt; 15 16 void add_edge(int u, int v, int w) { // from up to down 17 e[++edgeCnt] = Edge(v, G[u], w); 18 G[u] = edgeCnt; 19 } 20 21 int sz[MAXN], ch[MAXN], par[MAXN], dep[MAXN], top[MAXN]; 22 int id[MAXN], alloc; 23 int val[MAXN], cap[MAXN]; 24 25 void dfs_1(int u) { 26 sz[u] = 1, ch[u] = 0; 27 for (int i = G[u]; i; i = e[i].n) { 28 par[e[i].v] = u; 29 dep[e[i].v] = dep[u] + 1; 30 dfs_1(e[i].v); 31 sz[u] += sz[e[i].v]; 32 if (sz[e[i].v] > sz[ch[u]]) 33 ch[u] = e[i].v; 34 } 35 } 36 37 void dfs_2(int u, int anc) { 38 id[u] = ++alloc; 39 top[u] = anc; 40 if (ch[u]) 41 dfs_2(ch[u], anc); 42 for (int i = G[u]; i; i = e[i].n) if (e[i].v != ch[u]) { 43 dfs_2(e[i].v, e[i].v); 44 } 45 } 46 47 int mn[MAXN << 2], tag[MAXN << 2]; 48 int a[MAXN]; 49 int n, t; 50 51 inline void push_up(int o) { mn[o] = min(mn[o << 1], mn[o << 1 | 1]); } 52 inline void change(int o, int v) {mn[o] += v, tag[o] += v; } 53 void push_down(int o) { 54 if (tag[o]) { 55 change(o << 1, tag[o]); 56 change(o << 1 | 1, tag[o]); 57 tag[o] = 0; 58 } 59 } 60 61 void add(int ql, int qr, int x, int l=2, int r=n, int o=1) { 62 if (ql == l && qr == r) { 63 change(o, x); 64 return; 65 } 66 push_down(o); 67 int mid = (l + r) >> 1; 68 if (qr <= mid) 69 add(ql, qr, x, l, mid, o << 1); 70 else if (mid < ql) 71 add(ql, qr, x, mid + 1, r, o << 1 | 1); 72 else { 73 add(ql, mid, x, l, mid, o << 1); 74 add(mid + 1, qr, x, mid + 1, r, o << 1 | 1); 75 } 76 push_up(o); 77 } 78 79 int query(int ql, int qr, int l=2, int r=n, int o=1) { 80 if (ql == l && qr == r) 81 return mn[o]; 82 push_down(o); 83 int mid = (l + r) >> 1; 84 if (qr <= mid) 85 return query(ql, qr, l, mid, o << 1); 86 else if (mid < ql) 87 return query(ql, qr, mid + 1, r, o << 1 | 1); 88 else 89 return min(query(ql, mid, l, mid, o << 1), query(mid + 1, qr, mid + 1, r, o << 1 | 1)); 90 } 91 92 void build(int l, int r, int o=1) { 93 if (l == r) { 94 mn[o] = a[l]; 95 return; 96 } 97 int mid = (l + r) >> 1; 98 build(l, mid, o << 1); 99 build(mid + 1, r, o << 1 | 1); 100 push_up(o); 101 } 102 103 int qmin(int u) { 104 int ret = INT_MAX; 105 while (top[u] != 1) { 106 ret = min(ret, query(id[top[u]], id[u])); 107 u = par[top[u]]; 108 } 109 if (u == 1) 110 return ret; 111 return min(ret, query(id[ch[1]], id[u])); 112 } 113 114 void qadd(int u, int x) { 115 while (top[u] != 1) { 116 add(id[top[u]], id[u], x); 117 u = par[top[u]]; 118 } 119 if (u != 1) 120 add(id[ch[1]], id[u], x); 121 } 122 123 int main() { 124 ios::sync_with_stdio(false); 125 cin >> n >> t; 126 vector<pii> qwq; 127 for (int v = 2; v <= n + 1; ++v) { 128 int u, w; 129 cin >> val[v] >> u >> cap[v]; 130 u++; 131 add_edge(u, v, cap[v]); 132 qwq.push_back(pii(val[v], v)); 133 } 134 n++; 135 dfs_1(1); 136 dfs_2(1, 1); 137 for (int i = 1; i <= n; ++i) 138 a[id[i]] = cap[i]; 139 build(2, n); 140 sort(qwq.begin(), qwq.end(), greater<pii>()); 141 int64_t ans = 0; 142 for (auto& x : qwq) { 143 int mval = qmin(x.second); 144 if (mval < t) { 145 qadd(x.second, -mval); 146 ans += 1ll * mval * x.first; 147 t -= mval; 148 } else { 149 ans += 1ll * t * x.first; 150 break; 151 } 152 } 153 cout << ans << endl; 154 }
H.
码农题。令每次吃都吃有空缺的那一头，每次补都补满，这样可以保证吃到的永远最新鲜。设计状态然后输出状态转移即可。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 bool a[16]; 6 7 int encode() { 8 int ret = 0; 9 for (int i = 0; i < 16; ++i) 10 ret = ret * 2 + a[i]; 11 return ret; 12 } 13 14 void decode(int code) { 15 for (int i = 15; i >= 0; --i) { 16 a[i] = code % 2; 17 code >>= 1; 18 } 19 } 20 21 bool vis[(1 << 16) + 5]; 22 23 void print(int st1, int st2, string str) { 24 string l1, l2; 25 decode(st1); 26 for (int i = 0; i < 4; ++i) 27 l1 += (a[i] == 1 ? "*" : "."); 28 l1 += '|'; 29 for (int i = 4; i < 8; ++i) 30 l1 += (a[i] == 1 ? "*" : "."); 31 l1 += " " + str + " "; 32 33 for (int i = 8; i < 12; ++i) 34 l2 += (a[i] == 1 ? "*" : "."); 35 l2 += '|'; 36 for (int i = 12; i < 16; ++i) 37 l2 += (a[i] == 1 ? "*" : "."); 38 l2 += " "; 39 40 decode(st2); 41 for (int i = 0; i < 4; ++i) 42 l1 += (a[i] == 1 ? "*" : "."); 43 l1 += '|'; 44 for (int i = 4; i < 8; ++i) 45 l1 += (a[i] == 1 ? "*" : "."); 46 47 for (int i = 8; i < 12; ++i) 48 l2 += (a[i] == 1 ? "*" : "."); 49 l2 += '|'; 50 for (int i = 12; i < 16; ++i) 51 l2 += (a[i] == 1 ? "*" : "."); 52 53 cout << l1 << endl << l2 << "\n---\n"; 54 } 55 56 int count() { 57 int ret = 0; 58 for (int i = 0; i < 16; ++i) 59 ret += a[i]; 60 return ret; 61 } 62 63 inline int lfirst() { 64 for (int i = 0; i < 16; ++i) 65 if (a[i]) 66 return i; 67 return -1; 68 } 69 70 inline int rfirst() { 71 for (int i = 15; i >= 0; --i) 72 if (a[i]) 73 return i; 74 return -1; 75 } 76 77 void dfs(int st) { 78 if (vis[st]) 79 return; 80 vis[st] = true; 81 decode(st); 82 int cnt = count(); 83 if (cnt == 0) { 84 for (int i = 0; i < 10; ++i) 85 a[i] = 1; 86 print(st, encode(), "buy"); 87 dfs(encode()); 88 for (int i = 0; i < 10; ++i) 89 a[i] = 0; 90 } else if (cnt > 6) { 91 int x; 92 if (a[0] == 1) { 93 x = rfirst(); 94 } else if (a[15] == 1) { 95 x = lfirst(); 96 } 97 a[x] = 0; 98 print(st, encode(), "eat"); 99 dfs(encode()); 100 a[x] = 1; 101 } else if (cnt == 6) { 102 int x; 103 if (a[0] == 1) { 104 x = 0; 105 } else if (a[15] == 1) { 106 x = 15; 107 } 108 a[x] = 0; 109 print(st, encode(), "eat"); 110 dfs(encode()); 111 a[x] = 1; 112 } else if (cnt <= 5) { 113 int x = 0; 114 if (lfirst() == 10) { 115 x = rfirst(); 116 } else if (rfirst() == 5) { 117 x = lfirst(); 118 } 119 a[x] = 0; 120 print(st, encode(), "eat"); 121 dfs(encode()); 122 a[x] = 1; 123 124 if (lfirst() == 10) { 125 for (int i = 0; i < 10; ++i) { 126 a[i] = 1; 127 } 128 print(st, encode(), "buy"); 129 dfs(encode()); 130 for (int i = 0; i < 10; ++i) { 131 a[i] = 0; 132 } 133 } else if (rfirst() == 5) { 134 for (int i = 15; i > 5; --i) { 135 a[i] = 1; 136 } 137 print(st, encode(), "buy"); 138 dfs(encode()); 139 for (int i = 15; i > 5; --i) { 140 a[i] = 0; 141 } 142 } 143 } 144 } 145 146 int main() { 147 ios::sync_with_stdio(false); 148 dfs(0); 149 }