A - A+...+B Problem
可以取到的值一定是一段区间。所以答案即为max-min+1
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 int N, A, B;
37
38 int main()
39 {
40 giii(N, A, B);
41 B -= A;
42 int64 v = 1LL * (N - 1) * B - B + 1;
43 printf("%lld\n", v > 0 ? v : 0LL);
44 }
B - Evilator
首先如果方向不对就是2步，要不然就是一步，直接统计即可。
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 N = 1e5 + 10;
37
38 char str[N];
39
40 int n;
41
42 long long ans;
43
44 int main()
45 {
46 scanf("%s", str + 1);
47 n = strlen(str + 1);
48 for (int i = 1; i <= n; ++i)
49 {
50 if (str[i] == ‘U‘)
51 {
52 ans += 2 * (i - 1);
53 }
54 else
55 ans += i - 1;
56 }
57 //cerr << ans << endl;
58 reverse(str + 1, str + n + 1);
59 //cnt = 0;
60 for (int i = 1; i <= n; ++i)
61 {
62 if (str[i] == ‘D‘)
63 {
64 ans += 2 * (i - 1);
65 }
66 else
67 ans += i - 1;
68 }
69 printf("%lld\n", ans);
70 }
C - Nuske vs Phantom Thnook
图是一棵树，树的连通块个数=点数-边数，那么直接前缀和求点数、边数即可。
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 char str[2010][2010];
37
38 int a[2010][2010], b[2010][2010], c[2010][2010];
39
40 int n, m, q;
41
42 int main()
43 {
44 giii(n, m, q);
45 for (int i = 1; i <= n; ++i) scanf("%s", str[i] + 1);
46 for (int i = 1; i <= n; ++i)
47 {
48 for (int j = 1; j <= m; ++j)
49 if (str[i][j] == ‘1‘) ++a[i][j];
50 for (int j = 1; j < m; ++j)
51 if (str[i][j] == ‘1‘ && str[i][j + 1] == ‘1‘) ++b[i][j];
52 if (i < n)
53 for (int j = 1; j <= m; ++j)
54 if (str[i][j] == ‘1‘ && str[i + 1][j] == ‘1‘) ++c[i][j];
55 }
56 for (int i = 1; i <= n; ++i)
57 for (int j = 1; j <= m; ++j)
58 {
59 a[i][j] += a[i][j - 1] + a[i - 1][j] - a[i - 1][j - 1];
60 b[i][j] += b[i][j - 1] + b[i - 1][j] - b[i - 1][j - 1];
61 c[i][j] += c[i][j - 1] + c[i - 1][j] - c[i - 1][j - 1];
62 }
63 while (q--)
64 {
65 int x1, y1, x2, y2;
66 gii(x1, y1);
67 gii(x2, y2);
68 int A = a[x2][y2] + a[x1 - 1][y1 - 1] - a[x2][y1 - 1] - a[x1 - 1][y2];
69 //cerr << "A = " << a[x2][y2] << endl;
70 int B = b[x2][y2 - 1] + b[x1 - 1][y1 - 1] - b[x2][y1 - 1] - b[x1 - 1][y2 - 1];
71 int C = c[x2 - 1][y2] + c[x1 - 1][y1 - 1] - c[x2 - 1][y1 - 1] - c[x1 - 1][y2];
72 printf("%d\n", A - B - C);
73 }
74 return 0;
75 }
D - A or...or B Problem
我们考虑第一次出现不同位置的地方，令A和B最高位不同，然后分类讨论即可。
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 int64 A, B;
37
38 int main()
39 {
40 scanf("%lld%lld", &A, &B);
41 int p = 60;
42 if (A == B)
43 {
44 puts("1");
45 return 0;
46 }
47 while (!((A ^ B) & (1LL << p))) --p;
48 int q = p - 1;
49 while ((~q) && !(B & (1LL << q))) --q;
50 int64 x = (1LL << p) - (A & ((1LL << p) - 1));
51 int64 yl = (1LL << p), yr = (1LL << p) + (1LL << (q + 1));
52 int64 zl = (1LL << p) + (A & ((1LL << p) - 1)), zr = 2 * (1LL << p);
53 int64 ans = x + yr - yl + zr - zl;
54 if (zl <= yr)
55 ans -= yr - zl;
56 printf("%lld\n", ans);
57 return 0;
58 }
E - Mr.Aoki Incubator
我们发现本质就是区间覆盖问题，线段树dp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 const int N = 2e5 + 10;
39
40 int n;
41
42 PII t[N];
43
44 int id[N];
45
46 int L[N], R[N];
47
48 VI q[N];
49
50 bool comp(const int &i, const int &j)
51 {
52 return t[i].se < t[j].se;
53 }
54
55 int tr[N];
56
57 void add(int x, int v)
58 {
59 ++x;
60 for (; x < N; x += x & -x) (tr[x] += v) %= mod;
61 }
62
63 int sum(int x)
64 {
65 ++x;
66 int v = 0;
67 for (; x; x -= x & -x) (v += tr[x]) %= mod;
68 return v;
69 }
70
71 int query(int l, int r)
72 {
73 return (sum(r) - sum(l - 1) + mod) % mod;
74 }
75
76 int main()
77 {
78 gi(n);
79 for (int i = 1; i <= n; ++i)
80 gii(t[i].se, t[i].fi), id[i] = i;
81 sort(t + 1, t + n + 1);
82 sort(id + 1, id + n + 1, comp);
83 int mi = n, mx = 0;
84 for (int i = n; i; --i)
85 {
86 mi = min(mi, id[i]);
87 L[i] = mi;
88 }
89 for (int i = 1; i <= n; ++i)
90 {
91 mx = max(mx, id[i]);
92 R[i] = mx;
93 q[R[i]].pb(L[i]);
94 //cerr << L[i] << ", " << R[i] << endl;
95 }
96 add(0, 1);
97 for (int r = 1; r <= n; ++r)
98 for (auto l : q[r])
99 add(r, query(l - 1, r));
100 printf("%d\n", query(n, n));
101 return 0;
102 }
F - Kenus the Ancient Greek
只会最优解，目前还不会方案数
原文地址：https://www.cnblogs.com/AnzheWang/p/9696465.html
时间： 2024-10-08 08:05:14