A - 口算训练
题意：询问 $[L, R]$区间内 的所有数的乘积是否是D的倍数
思路：考虑分解质因数
显然，一个数$x > \sqrt{x} 的质因子只有一个$
那么我们考虑将小于$\sqrt {x}$ 的质因子用线段树维护
其他质因子用vector 维护存在性
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 100010
5 #define block 317
6 vector <int> fac[N], bigfac[10010];
7 int t, n, q;
8 int arr[N], pos[N], cnt;
9
10 void init()
11 {
12 cnt = 0;
13 memset(pos, 0, sizeof pos);
14 for (int i = 2; i < N; ++i)
15 {
16 if (pos[i] || i > block) continue;
17 for (int j = i * i; j < N; j += i)
18 pos[j] = 1;
19 }
20 for (int i = 2; i < N; ++i) if (!pos[i]) pos[i] = ++cnt;
21 for (int i = 2; i <= 100000; ++i)
22 {
23 int n = i;
24 for (int j = 2; j * j <= n; ++j)
25 {
26 while (n % j == 0)
27 {
28 n = n / j;
29 fac[i].push_back(j);
30 }
31 }
32 if (n != 1) fac[i].push_back(n);
33 sort(fac[i].begin(), fac[i].end());
34 }
35 }
36
37 int T[80];
38 struct SEG
39 {
40 #define M N * 300
41 int a[M], lson[M], rson[M], cnt;
42 void init() { cnt = 0; }
43 void pushup(int id) { a[id] = a[lson[id]] + a[rson[id]]; }
44 void update(int &id, int l, int r, int pos)
45 {
46 if (!id)
47 {
48 id = ++cnt;
49 a[id] = lson[id] = rson[id] = 0;
50 }
51 if (l == r)
52 {
53 ++a[id];
54 return;
55 }
56 int mid = (l + r) >> 1;
57 if (pos <= mid) update(lson[id], l, mid, pos);
58 else update(rson[id], mid + 1, r, pos);
59 pushup(id);
60 }
61 int query(int id, int l, int r, int ql, int qr)
62 {
63 if (!id) return 0;
64 if (l >= ql && r <= qr) return a[id];
65 int mid = (l + r) >> 1;
66 int res = 0;
67 if (ql <= mid) res += query(lson[id], l, mid, ql, qr);
68 if (qr > mid) res += query(rson[id], mid + 1, r, ql, qr);
69 return res;
70 }
71 }seg;
72
73 bool Get(int base, int need, int l, int r)
74 {
75 // printf("%d %d %d %d\n", base, need, l, r);
76 int have = 0;
77 if (base < block)
78 {
79 have = seg.query(T[pos[base]], 1, n, l, r);
80 if (have < need) return false;
81 }
82 else
83 {
84 have = upper_bound(bigfac[pos[base]].begin(), bigfac[pos[base]].end(), r) - lower_bound(bigfac[pos[base]].begin(), bigfac[pos[base]].end(), l);
85 if (have < need) return false;
86 }
87 return true;
88 }
89
90 bool work(int l, int r, int d)
91 {
92 if (d == 1) return true;
93 int base = fac[d][0], need = 1, have = 0;
94 for (int i = 1, len = fac[d].size(); i < len; ++i)
95 {
96 if (fac[d][i] != base)
97 {
98 if (Get(base, need, l, r) == false) return false;
99 base = fac[d][i];
100 need = 1;
101 }
102 else ++need;
103 }
104 return Get(base, need, l, r);
105 }
106
107 int main()
108 {
109 init();
110 scanf("%d", &t);
111 while (t--)
112 {
113 scanf("%d%d", &n, &q);
114 for (int i = 1; i < 10010; ++i) bigfac[i].clear();
115 for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
116 seg.init(); memset(T, 0, sizeof T);
117 for (int i = 1; i <= n; ++i)
118 {
119 for (auto it : fac[arr[i]])
120 {
121 if (it < block)
122 seg.update(T[pos[it]], 1, n, i);
123 else
124 bigfac[pos[it]].push_back(i);
125 }
126 }
127 for (int i = 1; i <= cnt; ++i) sort(bigfac[i].begin(), bigfac[i].end());
128 for (int i = 1, l, r, d; i <= q; ++i)
129 {
130 scanf("%d%d%d", &l, &r, &d);
131 puts(work(l, r, d) ? "Yes" : "No");
132 }
133 }
134 return 0;
135 }
B - 缺失的数据范围
题意：给出$a, b, k$ 求一个最大的n 使得 $n^{a} \cdot log ^b n <= k$
思路：二分 但是要注意用高精度，求log的时候 平方逼近
1 import java.io.BufferedInputStream;
2 import java.util.Scanner;
3 import java.math.*;
4
5 public class Main {
6
7 public static BigInteger Get(BigInteger x, int a, int b)
8 {
9 BigInteger two = BigInteger.valueOf(2);
10 BigInteger tmp = BigInteger.ONE;
11 BigInteger cnt = BigInteger.ZERO;
12 while (tmp.compareTo(x) < 0)
13 {
14 tmp = tmp.multiply(two);
15 cnt = cnt.add(BigInteger.ONE);
16 }
17 BigInteger res = x.pow(a).multiply(cnt.pow(b));
18 return res;
19 }
20
21
22 public static void main(String[] args) {
23 Scanner in = new Scanner(new BufferedInputStream(System.in));
24 int t = in.nextInt();
25 int a, b; BigInteger k, l, r, mid, ans, zero, two, one;
26 zero = BigInteger.ZERO;
27 two = BigInteger.valueOf(2);
28 one = BigInteger.valueOf(1);
29 while (t-- != 0)
30 {
31 a = in.nextInt();
32 b = in.nextInt();
33 k = in.nextBigInteger();
34 l = one;
35 r = k;
36 ans = zero;
37 while (r.subtract(l).compareTo(zero) >= 0)
38 {
39 mid = l.add(r).divide(two);
40 if (Get(mid, a, b).compareTo(k) <= 0)
41 {
42 ans = mid;
43 l = mid.add(one);
44 }
45 else
46 r = mid.subtract(one);
47 }
48 System.out.println(ans);
49 }
50 in.close();
51 }
52 }
C - 寻宝游戏
留坑。
D - 奢侈的旅行
留坑。
E - 对称数
留坑。
F - 赛题分析
水。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1010
5 #define INF 0x3f3f3f3f
6 int t, n, m;
7 int arr[N], brr[N];
8
9 int main()
10 {
11 scanf("%d", &t);
12 for (int kase = 1; kase <= t; ++kase)
13 {
14 printf("Problem %d:\n", kase + 1000);
15 scanf("%d%d", &n, &m);
16 for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
17 if (n == 0) puts("Shortest judge solution: N/A bytes.");
18 else printf("Shortest judge solution: %d bytes.\n", *min_element(arr + 1, arr + 1 + n));
19 for (int i = 1; i <= m; ++i) scanf("%d", brr + i);
20 if (m == 0) puts("Shortest team solution: N/A bytes.");
21 else printf("Shortest team solution: %d bytes.\n", *min_element(brr + 1, brr + 1 + m));
22 }
23 return 0;
24 }
G - quailty算法
留坑。
H - SA-IS后缀数组
题意：求每对相邻的后缀字符的字典序大小
思路：从后面往前推，只考虑最高位，如果最高位相同，那么问题转化为一个子问题
比如 cm 和 acm 考虑最高位 如果最高位相同 问题转化为 m 和 cm 的大小关系
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1000010
5 int t, n;
6 char s[N];
7 int ans[N];
8
9 int main()
10 {
11 scanf("%d", &t);
12 while (t--)
13 {
14 int n;
15 scanf("%d%s", &n, s);
16 int len = strlen(s);
17 if (s[len - 1] != s[len - 2]) ans[len - 2] = s[len - 2] > s[len - 1];
18 else ans[len - 2] = 1;
19 for (int i = len - 3; i >= 0; --i)
20 {
21 if (s[i] != s[i + 1]) ans[i] = s[i] > s[i + 1];
22 else ans[i] = ans[i + 1];
23 }
24 for (int i = 0; i < len - 1; ++i) printf("%c", ans[i] ? ‘>‘ : ‘<‘);
25 puts("");
26 }
27 return 0;
28 }
I - 回文树
留坑。
J - 代码派对
留坑。
K - CCPC直播
按题意模拟即可。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 int t;
5 char s[20], sta[20];
6 int Rank, x, pro;
7
8 int main()
9 {
10 scanf("%d", &t);
11 while (t--)
12 {
13 scanf("%d%s%d%s", &Rank, s, &pro, sta);
14 printf("%3d|%-16s|%d|[", Rank, s, pro);
15 if (strcmp(sta, "Running") == 0)
16 {
17 scanf("%d", &x);
18 for (int i = 0; i < x; ++i) putchar(‘X‘);
19 for (int i = x; i < 10; ++i) putchar(‘ ‘);
20 printf("]\n");
21 }
22 else
23 {
24 if (strcmp(sta, "FB") == 0) strcpy(sta, "AC*");
25 printf(" ");
26 printf("%-6s]\n", sta);
27 }
28 }
29 return 0;
30 }
原文地址：https://www.cnblogs.com/Dup4/p/9851609.html
时间： 2024-10-11 16:30:52