A - Sorted Arrays
贪心，看看不下降和不上升最长能到哪，直接转移过去即可。
1 //waz
2 #include <bits/stdc++.h>
3
4 using namespace std;
5
6 #define mp make_pair
7 #define pb push_back
8 #define fi first
9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22
23 int F()
24 {
25 char ch;
26 int x, a;
27 while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
28 if (ch == ‘-‘) ch = getchar(), a = -1;
29 else a = 1;
30 x = ch - ‘0‘;
31 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
32 x = (x << 1) + (x << 3) + ch - ‘0‘;
33 return a * x;
34 }
35
36 const int mod = 1e9 + 7;
37
38 int n, a[100010], cnt;
39
40 int main()
41 {
42 gi(n);
43 for (int i = 1; i <= n; ++i) gi(a[i]);
44 for (int i = 1; i <= n; ++i)
45 {
46 int j = i, k = i;
47 while (j < n && a[j + 1] >= a[j]) ++j;
48 while (k < n && a[k + 1] <= a[k]) ++k;
49 i = max(j, k);
50 ++cnt;
51 }
52 printf("%d\n", cnt);
53 return 0;
54 }
B - Hamiltonish Path
随便选一个边，从两头开始dfs到不能走就就好了。
1 //waz
2 #include <bits/stdc++.h>
3
4 using namespace std;
5
6 #define mp make_pair
7 #define pb push_back
8 #define fi first
9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22
23 int F()
24 {
25 char ch;
26 int x, a;
27 while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
28 if (ch == ‘-‘) ch = getchar(), a = -1;
29 else a = 1;
30 x = ch - ‘0‘;
31 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
32 x = (x << 1) + (x << 3) + ch - ‘0‘;
33 return a * x;
34 }
35
36 const int N = 1e5 + 10;
37
38 int n, m;
39
40 VI edge[N];
41
42 VI l, r;
43
44 bool vis[N];
45
46 int main()
47 {
48 gii(n, m);
49 for (int i = 1; i <= m; ++i)
50 {
51 int u, v;
52 gii(u, v);
53 edge[u].pb(v);
54 edge[v].pb(u);
55 if (i == 1) vis[u] = vis[v] = 1, l.pb(u), r.pb(v);
56 }
57 while (1)
58 {
59 int x = -1;
60 for (auto v : edge[l.back()])
61 if (!vis[v]) x = v;
62 if (x == -1) break;
63 l.pb(x); vis[x] = 1;
64 }
65 while (1)
66 {
67 int x = -1;
68 for (auto v : edge[r.back()])
69 if (!vis[v]) x = v;
70 if (x == -1) break;
71 r.pb(x); vis[x] = 1;
72 }
73 printf("%d\n", SZ(l) + SZ(r));
74 reverse(ALL(l));
75 for (auto x : l)
76 printf("%d ", x);
77 for (auto x : r)
78 printf("%d ", x);
79 return 0;
80 }
C - Ants on a Circle
很明显的套路，蚂蚁的相对位置肯定不变，那么我们算出每个蚂蚁走了多少，最后排序一下，主要是算第一个位置是谁，再统计一下经过了圆多少次就好了。
1 //waz
2 #include <bits/stdc++.h>
3
4 using namespace std;
5
6 #define mp make_pair
7 #define pb push_back
8 #define fi first
9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22
23 int F()
24 {
25 char ch;
26 int x, a;
27 while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
28 if (ch == ‘-‘) ch = getchar(), a = -1;
29 else a = 1;
30 x = ch - ‘0‘;
31 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
32 x = (x << 1) + (x << 3) + ch - ‘0‘;
33 return a * x;
34 }
35
36 const int N = 1e5 + 10;
37
38 int n, l, t;
39
40 int x[N], w[N];
41
42 int main()
43 {
44 giii(n, l, t);
45 long long v = 0;
46 for (int i = 1; i <= n; ++i)
47 {
48 gii(x[i], w[i]);
49 if (w[i] == 1)
50 {
51 x[i] += t;
52 v += x[i] / l;
53 x[i] %= l;
54 }
55 else
56 {
57 x[i] -= t;
58 if (x[i] < 0)
59 {
60 v += (x[i] - (l - 1)) / l;
61 }
62 x[i] %= l;
63 if (x[i] < 0) x[i] += l;
64 }
65 }
66 sort(x + 1, x + n + 1);
67 v %= n;
68 if (v < 0) v += n;
69 for (int i = 1; i <= n; ++i)
70 {
71 int j = i + v;
72 if (j > n) j -= n;
73 printf("%d\n", x[j]);
74 }
75 return 0;
76 }
D - Piling Up
直接f[i][j]dp会算重，我们加一维0/1表示是否曾经到达过0，这样子就不会重了。
1 //waz
2 #include <bits/stdc++.h>
3
4 using namespace std;
5
6 #define mp make_pair
7 #define pb push_back
8 #define fi first
9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22
23 int F()
24 {
25 char ch;
26 int x, a;
27 while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
28 if (ch == ‘-‘) ch = getchar(), a = -1;
29 else a = 1;
30 x = ch - ‘0‘;
31 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
32 x = (x << 1) + (x << 3) + ch - ‘0‘;
33 return a * x;
34 }
35
36 const int N = 3010, mod = 1e9 + 7;
37
38 int inc(int a, int b) { a += b; return a >= mod ? a - mod : a; }
39
40 int n, m, f[N][N][2];
41
42 int main()
43 {
44 gii(n, m);
45 for (int i = 0; i <= n; ++i) f[0][i][!!i] = 1;
46 for (int i = 1; i <= m; ++i)
47 {
48 for (int j = 0; j <= n; ++j)
49 {
50 //2 * x
51 f[i][j][0] = inc(f[i][j][0], f[i - 1][j + 1][0]);
52 f[i][j][!!j] = inc(f[i][j][!!j], f[i - 1][j + 1][1]);
53
54 //x ... y
55 if (j) f[i][j][0] = inc(f[i][j][0], f[i - 1][j][0]);
56 if (j) f[i][j][!!(j - 1)] = inc(f[i][j][!!(j - 1)], f[i - 1][j][1]);
57
58 //y ... x
59 if (j != n) f[i][j][0] = inc(f[i][j][0], f[i - 1][j][0]);
60 if (j != n) f[i][j][!!j] = inc(f[i][j][!!j], f[i - 1][j][1]);
61
62 //2 * y
63 if (j) f[i][j][0] = inc(f[i][j][0], f[i - 1][j - 1][0]);
64 if (j) f[i][j][1] = inc(f[i][j][1], f[i - 1][j - 1][1]);
65 }
66 }
67 int ans = 0;
68 for (int i = 0; i <= n; ++i) ans = inc(ans, f[m][i][0]);
69 printf("%d\n", ans);
70 return 0;
71 }
E - Placing Squares
写出n^2的dp式子:f[i]=sigma(f[j]*(i-j)^2)，对于所有x[i]，f[x[i]]=0。
我们考虑式子的组合意义，就是两个不同的球放在(i-j)箱子的方案数。
我们就可以dp[i][0/1/2]来线性递推了。
就可以用矩阵优化了。
1 //waz
2 #include <bits/stdc++.h>
3
4 using namespace std;
5
6 #define mp make_pair
7 #define pb push_back
8 #define fi first
9 #define se second
10 #define ALL(x) (x).begin(), (x).end()
11 #define SZ(x) ((int)((x).size()))
12
13 typedef pair<int, int> PII;
14 typedef vector<int> VI;
15 typedef long long int64;
16 typedef unsigned int uint;
17 typedef unsigned long long uint64;
18
19 #define gi(x) ((x) = F())
20 #define gii(x, y) (gi(x), gi(y))
21 #define giii(x, y, z) (gii(x, y), gi(z))
22
23 int F()
24 {
25 char ch;
26 int x, a;
27 while (ch = getchar(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘);
28 if (ch == ‘-‘) ch = getchar(), a = -1;
29 else a = 1;
30 x = ch - ‘0‘;
31 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘)
32 x = (x << 1) + (x << 3) + ch - ‘0‘;
33 return a * x;
34 }
35
36 const int mod = 1e9 + 7;
37
38 int inc(int a, int b) { a += b; return a >= mod ? a - mod : a; }
39
40 int n, m, x[100010];
41
42 void task1()
43 {
44 static int dp[100010][3];
45 dp[0][0] = 1;
46 static map<int, bool> X;
47 for (int i = 1; i <= m; ++i) X[x[i]] = 1;
48 for (int i = 1; i <= n; ++i)
49 {
50 dp[i][0] = dp[i - 1][0];
51 dp[i][1] = inc(dp[i - 1][0], dp[i - 1][1]);
52 dp[i][2] = inc(inc(inc(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][1]), dp[i - 1][2]);
53 if(!X.count(i - 1))
54 {
55 dp[i][0] = inc(dp[i][0], dp[i - 1][2]);
56 dp[i][1] = inc(dp[i][1], dp[i - 1][2]);
57 dp[i][2] = inc(dp[i][2], dp[i - 1][2]);
58 }
59 }
60 printf("%d\n", dp[n][2]);
61 }
62
63 struct matrix
64 {
65 int a[3][3];
66 friend matrix operator * (matrix a, matrix b)
67 {
68 matrix c;
69 memset(c.a, 0, sizeof c.a);
70 for (int k = 0; k < 3; ++k)
71 for (int i = 0; i < 3; ++i)
72 for (int j = 0; j < 3; ++j)
73 c.a[i][j] = (c.a[i][j] + 1LL * a.a[i][k] * b.a[k][j]) % mod;
74 return c;
75 }
76 } one, A, B, C;
77
78 matrix fpow(matrix a, int x)
79 {
80 matrix ret = one;
81 for (; x; x >>= 1)
82 {
83 if (x & 1) ret = ret * a;
84 a = a * a;
85 }
86 return ret;
87 }
88
89 void work()
90 {
91 /*
92 1,0,1
93 1,1,1
94 1,2,2
95 */
96 /*
97 1,0,0
98 1,1,0
99 1,2,1
100 */
101 one.a[0][0] = one.a[1][1] = one.a[2][2] = 1;
102 A.a[0][0] = 1, A.a[0][1] = 0, A.a[0][2] = 1;
103 A.a[1][0] = 1, A.a[1][1] = 1, A.a[1][2] = 1;
104 A.a[2][0] = 1, A.a[2][1] = 2, A.a[2][2] = 2;
105 B.a[0][0] = 1, B.a[0][1] = 0, B.a[0][2] = 0;
106 B.a[1][0] = 1, B.a[1][1] = 1, B.a[1][2] = 0;
107 B.a[2][0] = 1, B.a[2][1] = 2, B.a[2][2] = 1;
108 matrix ret = one;
109 for (int i = 1; i <= m; ++i)
110 {
111 ret = ret * fpow(A, x[i] - x[i - 1] - 1);
112 ret = ret * B;
113 }
114 ret = ret * fpow(A, n - x[m]);
115 C.a[0][0] = 1;
116 ret = ret * C;
117 printf("%d\n", ret.a[2][0]);
118 }
119
120 int main()
121 {
122 gii(n, m);
123 for (int i = 1; i <= m; ++i) gi(x[i]);
124 //task1();
125 work();
126 return 0;
127 }
F - Two Faced Cards
不会做，先咕着（目前已经咕了4题了qwq）。
原文地址：https://www.cnblogs.com/AnzheWang/p/9678703.html
时间： 2024-10-08 08:05:16