1. All X
1.1 基本思路
k和c的范围都不大，因此可以考虑迭代找循环节，然后求余数，判定是否相等。这题还是挺简单的。
1.2 代码
1 /* 5690 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 typedef long long LL;
45 const int maxn = 10005;
46 int x;
47 LL m;
48 int c, k;
49 int visit[maxn];
50
51 int main() {
52 ios::sync_with_stdio(false);
53 #ifndef ONLINE_JUDGE
54 freopen("data.in", "r", stdin);
55 freopen("data.out", "w", stdout);
56 #endif
57
58 int t;
59 int tmp, ntmp;
60 LL i, cycle;
61
62 scanf("%d", &t);
63 rep(tt, 1, t+1) {
64 scanf("%d%I64d%d%d", &x,&m,&k,&c);
65 printf("Case #%d:\n", tt);
66 memset(visit, -1, sizeof(visit));
67 ntmp = x % k;
68 cycle = -1;
69 for (i=1; i<=m; ++i) {
70 if (visit[tmp=ntmp] != -1) {
71 cycle = i - visit[tmp];
72 break;
73 }
74 visit[tmp] = i;
75 ntmp = (10*tmp + x) % k;
76 }
77
78 if (cycle == -1) {
79 puts(tmp==c ? "Yes":"No");
80 } else {
81 LL n = ((m - visit[tmp]) % cycle + cycle) % cycle;
82 for (i=0; i<n; ++i)
83 tmp = (10*tmp + x) % k;
84 puts(tmp==c ? "Yes":"No");
85 }
86 }
87
88 #ifndef ONLINE_JUDGE
89 printf("time = %ldms.\n", clock());
90 #endif
91
92 return 0;
93 }
2. Sitting in Line
2.1 基本思路
N的范围很小，刚开始以为这是一个数学题，后来发现状态DP可解。
$dp[st][n]$st表示当前状态（即当前已经加入数字），n表示当前排列的最末端数字。显然有状态转移方程
\[
dp[nst][k] = \max(dp[nst][k], dp[st|(1<<k)][j]+a_j \times a_k), \\
\qquad \text{其中 } st\&(1<<k) == 0\text{ 并且 }pos_k==-1 || pos_k==getBits(st)
\]
2.2 代码
1 /* 5691 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 typedef long long LL;
45 LL INF = 0x3f3f3f3f3f3f3f3f;
46 LL NEG_INF = 0xc0c0c0c0c0c0c0c0;
47 const int maxn = 16;
48 LL dp[1<<maxn][maxn];
49 int a[maxn], p[maxn];
50 int Bits[1<<maxn];
51 int n;
52
53 int getBits(int x) {
54 int ret = 0;
55
56 while (x) {
57 ret += x & 1;
58 x >>= 1;
59 }
60
61 return ret;
62 }
63
64 void init() {
65 const int mst = 1<<16;
66 rep(i, 0, mst) Bits[i] = getBits(i);
67 }
68
69 void solve() {
70 memset(dp, 0xC0, sizeof(dp));
71 rep(i, 0, n) {
72 if (p[i]==-1 || p[i]==0)
73 dp[1<<i][i] = 0;
74 }
75
76 const int mst = 1<<n;
77 #ifndef ONLINE_JUDGE
78 printf("NEG_INF = %I64d\n", NEG_INF);
79 #endif
80 rep(i, 0, mst) {
81 const int idx = Bits[i];
82 rep(j, 0, n) {
83 if (dp[i][j] == NEG_INF) continue;
84 rep(k, 0, n) {
85 if (i & (1<<k)) continue;
86 if (p[k]==-1 || p[k]==idx) {
87 int nst = i | 1<<k;
88 dp[nst][k] = max(dp[nst][k], dp[i][j]+a[j]*a[k]);
89 }
90 }
91 }
92 }
93
94 LL ans = NEG_INF;
95 rep(j, 0, n)
96 ans = max(ans, dp[mst-1][j]);
97 printf("%I64d\n", ans);
98 }
99
100 int main() {
101 ios::sync_with_stdio(false);
102 #ifndef ONLINE_JUDGE
103 freopen("data.in", "r", stdin);
104 freopen("data.out", "w", stdout);
105 #endif
106
107 int t;
108
109 init();
110 scanf("%d", &t);
111 rep(tt, 1, t+1) {
112 scanf("%d", &n);
113 rep(i, 0, n)
114 scanf("%d%d", &a[i],&p[i]);
115 printf("Case #%d:\n", tt);
116 solve();
117 }
118
119 #ifndef ONLINE_JUDGE
120 printf("time = %ldms.\n", clock());
121 #endif
122
123 return 0;
124 }
3. Snacks
3.1 基本思路
这个题刚开始看以为是树链剖分，巨难无比，赛后发现就是个树形结构转线性结构，然后使用线段树维护最大值就好了，使用下lazy标记直接A了。
3.2 代码
1 /* 5692 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 #pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 struct edge_t {
45 int v, nxt;
46 };
47
48 typedef long long LL;
49 const int maxn = 1e5+5;
50 const int maxv = maxn;
51 const int maxe = maxn * 2;
52 int head[maxv], l, cnt;
53 edge_t E[maxe];
54 LL mx[maxn<<2], delta[maxn<<2];
55 int L[maxn], R[maxn], a[maxn];
56 int LLL, RR, w;
57 int n, m;
58
59 void init() {
60 memset(head, -1, sizeof(head));
61 l = cnt = 0;
62 memset(mx, 0, sizeof(mx));
63 memset(delta, 0, sizeof(delta));
64 }
65
66 void addEdge(int u, int v) {
67 E[l].v = v;
68 E[l].nxt = head[u];
69 head[u] = l++;
70
71 E[l].v = u;
72 E[l].nxt = head[v];
73 head[v] = l++;
74 }
75
76 void dfs(int u, int fa) {
77 L[u] = ++cnt;
78 for (int k=head[u]; k!=-1; k=E[k].nxt) {
79 const int& v = E[k].v;
80 if (v == fa) continue;
81 dfs(v, u);
82 }
83 R[u] = cnt;
84 }
85
86 inline void PushDown(int rt) {
87 if (delta[rt]) {
88 int lb = rt<<1, rb = lb | 1;
89 delta[lb] += delta[rt];
90 delta[rb] += delta[rt];
91 mx[lb] += delta[rt];
92 mx[rb] += delta[rt];
93 delta[rt] = 0;
94 }
95 }
96
97 inline void PushUp(int rt) {
98 mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
99 }
100
101 void Update(int l, int r, int rt) {
102 if (LLL<=l && RR>=r) {
103 mx[rt] += w;
104 delta[rt] += w;
105 return ;
106 }
107
108 PushDown(rt);
109 int mid = (l + r) >> 1;
110
111 if (RR <= mid) {
112 Update(lson);
113 } else if (LLL > mid) {
114 Update(rson);
115 } else {
116 Update(lson);
117 Update(rson);
118 }
119
120 PushUp(rt);
121 }
122
123 LL Query(int l, int r, int rt) {
124 if (LLL<=l && RR>=r) {
125 return mx[rt];
126 }
127
128 PushDown(rt);
129 int mid = (l + r) >> 1;
130
131 if (RR <= mid)
132 return Query(lson);
133 else if (LLL > mid)
134 return Query(rson);
135 else
136 return max(Query(lson), Query(rson));
137 }
138
139 void solve() {
140 dfs(0, -1);
141
142 rep(i, 0, n) {
143 scanf("%d", &a[i]);
144 LLL = L[i];
145 RR = R[i];
146 w = a[i];
147 Update(1, n, 1);
148 }
149
150 int op, x, y;
151 LL ans;
152
153 rep(i, 0, m) {
154 scanf("%d%d", &op, &x);
155 LLL = L[x];
156 RR = R[x];
157 if (op) {
158 ans = Query(1, n, 1);
159 printf("%I64d\n", ans);
160 } else {
161 scanf("%d", &y);
162 y -= a[x];
163 w = y;
164 Update(1, n, 1);
165 a[x] += y;
166 }
167 }
168 }
169
170 int main() {
171 ios::sync_with_stdio(false);
172 #ifndef ONLINE_JUDGE
173 freopen("data.in", "r", stdin);
174 freopen("data.out", "w", stdout);
175 #endif
176
177 int t;
178 int u, v;
179
180 scanf("%d", &t);
181 rep(tt, 1, t+1) {
182 scanf("%d%d", &n,&m);
183 init();
184 rep(i, 1, n) {
185 scanf("%d%d", &u,&v);
186 addEdge(u, v);
187 }
188 printf("Case #%d:\n", tt);
189 solve();
190 }
191
192 #ifndef ONLINE_JUDGE
193 printf("time = %ldms.\n", clock());
194 #endif
195
196 return 0;
197 }
4. D Game
4.1 基本思路
$n,m \in [1,300]$这数据范围很小，基本思路是DP.
不妨令$dp[i][j]$表示从第$i$个数字到第$j$个数字能否通过全部被删掉，$mx[i]$表示前$i$个数字中可以最多可以删掉的数字。显然有动态转移方程
\[
mx[i] = \max(mx[i], mx[j-1]+i-j+1), \text{ if } dp[j][i]=True
\]
因此，紧急考虑$dp[j][i]$的情况即可：
(1) $dp[j][k] \& dp[k+1][i], k \in [j+1, i)$
(2) $dp[j+1][i-1], (a_i-a_j) \in D$
(3) $dp[j+1][k-1] \& dp[k+1][i-1], (a_i-a_j)/2 \in D, a_k*2 = a_i+a_j, k \in [j+1, i)$
4.2 代码
1 /* 5693 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 const int maxn = 305;
45 bool dp[maxn][maxn];
46 int mx[maxn];
47 int a[maxn];
48 int n, m;
49 sti st;
50
51 void solve() {
52 memset(dp, 0, sizeof(dp));
53
54 rep(i, 1, n+1) {
55 dp[i][i] = false;
56 dp[i][i-1] = true;
57 per(j, 1, i) {
58 rep(k, j+1, i)
59 dp[j][i] |= dp[j][k] & dp[k+1][i];
60 if (st.count(a[i]-a[j])) {
61 dp[j][i] |= dp[j+1][i-1];
62 }
63 if ((a[i]-a[j])%2==0 && st.count((a[i]-a[j])>>1)) {
64 rep(k, j+1, i) {
65 if (a[k]+a[k] == a[j]+a[i]) {
66 dp[j][i] |= dp[j+1][k-1] & dp[k+1][i-1];
67 }
68 }
69 }
70 }
71 }
72
73 mx[0] = 0;
74 rep(i, 1, n+1) {
75 mx[i] = mx[i-1];
76 rep(j, 1, i) {
77 if (dp[j][i])
78 mx[i] = max(mx[i], mx[j-1]+i-j+1);
79 }
80 }
81
82 printf("%d\n", mx[n]);
83 }
84
85 int main() {
86 ios::sync_with_stdio(false);
87 #ifndef ONLINE_JUDGE
88 freopen("data.in", "r", stdin);
89 freopen("data.out", "w", stdout);
90 #endif
91
92 int t;
93 int x;
94
95 scanf("%d", &t);
96 while (t--) {
97 scanf("%d%d", &n,&m);
98 rep(i, 1, n+1) scanf("%d", a+i);
99 st.clr();
100 rep(i, 0, m) {
101 scanf("%d", &x);
102 st.insert(x);
103 }
104 solve();
105 }
106
107 #ifndef ONLINE_JUDGE
108 printf("time = %ldms.\n", clock());
109 #endif
110
111 return 0;
112 }
5. BD String
5.1 基本思路
不妨令$f(n)$表示$S(n)$字符串的长度，显然很容易得到$f(n) = 2^n-1$。因此，最终串的长度为$f(2^{1000}) = 2^{1001}-1$，而$L,R \in [1,10^{18}]$。
显然原问题等价于计算$S(60)$的$[L,R]$范围内的B的个数。
算法很简单，就是个递归。可以考虑$S(k)$的左半部分和右半部分，递归时，加上剪枝$R-L+1==f(dep)$即可，此时B的数量为$f(dep-1)$,D的数量为$f(dep-1)-1$。
5.2 代码
1 /* 5694 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 #pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 typedef long long LL;
45 LL Base[62];
46 LL L, R;
47
48 void init() {
49 rep(i, 0, 62)
50 Base[i] = 1LL << i;
51 }
52
53 LL dfsD(LL, LL, int);
54 LL dfsB(LL, LL, int);
55
56 LL dfsD(LL L, LL R, int dep) {
57 if (L > R) return 0;
58 if (dep == 0) return 0;
59 if (dep == 1) return 0;
60
61 if (R-L+1 == Base[dep]-1) {
62 return Base[dep-1]-1;
63 }
64 LL mid = Base[dep-1];
65 if (R < mid) {
66 return dfsD(L, R, dep-1);
67 } else if (L > mid) {
68 LL nxtL = mid - 1 - (R - mid) + 1;
69 LL nxtR = mid - 1 - (L - mid) + 1;
70 return dfsB(nxtL, nxtR, dep-1);
71
72 } else {
73 LL lret = dfsD(L, mid-1, dep-1);
74 LL rret = 0;
75
76 if (R > mid) {
77 LL nxtL = mid-1 - (R-mid) + 1;
78 LL nxtR = mid-1;
79 rret = dfsB(nxtL, nxtR, dep-1);
80 }
81
82 return lret + rret;
83 }
84 }
85
86 LL dfsB(LL L, LL R, int dep) {
87 if (L > R) return 0;
88 if (dep == 0) return 0;
89 if (dep == 1) return 1;
90
91 if (R-L+1 == Base[dep]-1) {
92 return Base[dep-1];
93 }
94 LL mid = Base[dep-1];
95 if (R < mid) {
96 return dfsB(L, R, dep-1);
97
98 } else if (L > mid) {
99 LL nxtL = mid - 1 - (R - mid) + 1;
100 LL nxtR = mid - 1 - (L - mid) + 1;
101 return dfsD(nxtL, nxtR, dep-1);
102
103 } else {
104 LL lret = dfsB(L, mid-1, dep-1);
105 LL rret = 0;
106
107 if (R > mid) {
108 LL nxtL = mid-1 - (R-mid) + 1;
109 LL nxtR = mid-1;
110 rret = dfsD(nxtL, nxtR, dep-1);
111 }
112
113 return lret + 1 + rret;
114 }
115 }
116
117 void solve() {
118 LL ans = dfsB(L, R, 60);
119
120 printf("%I64d\n", ans);
121 }
122
123 int main() {
124 ios::sync_with_stdio(false);
125 #ifndef ONLINE_JUDGE
126 freopen("data.in", "r", stdin);
127 freopen("data.out", "w", stdout);
128 #endif
129
130 int t;
131
132 init();
133 scanf("%d", &t);
134 while (t--) {
135 scanf("%I64d%I64d", &L, &R);
136 solve();
137 }
138
139 #ifndef ONLINE_JUDGE
140 printf("time = %ldms.\n", clock());
141 #endif
142
143 return 0;
144 }
6. Gym Class
6.1 基本思路
水题，根据依赖性建图，然后拓扑排序就好了。注意拓扑的时候使用优先队列可求最值。
6.2 代码
1 /* */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 #pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 struct edge_t {
45 int v, nxt;
46 };
47
48 typedef long long LL;
49 const int maxn = 1e5+5;
50 const int maxv = maxn;
51 const int maxe = maxn;
52 int head[maxv], l;
53 edge_t E[maxe];
54 int deg[maxv];
55 bool visit[maxv];
56 int n, m;
57
58 void init() {
59 memset(head, -1, sizeof(head));
60 memset(deg, 0, sizeof(deg));
61 memset(visit, false, sizeof(visit));
62 l = 0;
63 }
64
65 inline void addEdge(int u, int v) {
66 ++deg[v];
67 E[l].v = v;
68 E[l].nxt = head[u];
69 head[u] = l++;
70 }
71
72 void solve() {
73 priority_queue<int> Q;
74 LL ans = 0;
75 int u, v, k, mn = INT_MAX;
76
77 rep(i, 1, n+1) {
78 if (deg[i] == 0) {
79 Q.push(i);
80 visit[i] = true;
81 }
82 }
83
84 while (!Q.empty()) {
85 u = Q.top();
86 Q.pop();
87 mn = min(u, mn);
88 ans += mn;
89 for (k=head[u]; k!=-1; k=E[k].nxt) {
90 v = E[k].v;
91 if (!visit[v] && --deg[v]==0) {
92 Q.push(v);
93 visit[v] = true;
94 }
95 }
96 }
97
98 printf("%I64d\n", ans);
99 }
100
101 int main() {
102 ios::sync_with_stdio(false);
103 #ifndef ONLINE_JUDGE
104 freopen("data.in", "r", stdin);
105 freopen("data.out", "w", stdout);
106 #endif
107
108 int t;
109 int u, v;
110
111 scanf("%d", &t);
112 while (t--) {
113 scanf("%d%d", &n, &m);
114 init();
115 rep(i, 0, m) {
116 scanf("%d%d", &u,&v);
117 addEdge(u, v);
118 }
119 solve();
120 }
121
122 #ifndef ONLINE_JUDGE
123 printf("time = %ldms.\n", clock());
124 #endif
125
126 return 0;
127 }
时间： 2024-10-24 09:52:43