"字节跳动杯"2018中国大学生程序设计竞赛-女生专场 Solution

Posted 2021-01-15 dup4

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
", 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 }

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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 }

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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:
", 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.
", *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.
", *min_element(brr + 1, brr + 1 + m));
22     }
23     return 0;
24 }

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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 }

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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("]
");
21         }
22         else
23         {
24             if (strcmp(sta, "FB") == 0) strcpy(sta, "AC*");
25             printf("    ");
26             printf("%-6s]
", sta);
27         }
28     }
29     return 0;
30 }

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