ZOJ Monthly, March 2018 Solution

Posted 2021-01-14 dup4

A - Easy Number Game

水。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 100010
 6 ll arr[N];
 7 int n, m;
 8 
 9 int main()
10 {
11     int t; scanf("%d", &t);
12     while (t--)
13     {
14         scanf("%d%d", &n, &m);
15         for (int i = 1; i <= n; ++i) scanf("%lld", arr + i);
16         sort(arr + 1, arr + 1 + n);
17         ll res = 0;
18         for (int i = 1; i <= m; ++i)
19             res += arr[i] * arr[2 * m - i + 1]; 
20         printf("%lld
", res);
21     }
22     return 0;
23 }

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B - Lucky Man

题意：判断大数开根后的奇偶性

思路：牛顿迭代法

 1 import java.io.BufferedInputStream;
 2 import java.util.Scanner;
 3 import java.math.*;
 4 
 5 public class Main {
 6 
 7     public static void main(String[] args) {
 8         Scanner in = new Scanner(new BufferedInputStream(System.in));
 9         int t = in.nextInt();
10         BigInteger a, x, two; String n;
11         two = BigInteger.valueOf(2);
12         while (t-- != 0)
13         {
14             n = in.next();
15             a = new BigInteger(n);
16             x = new BigInteger(n.substring(0, n.length() / 2 + 1));
17             while (a.compareTo(x.multiply(x)) < 0)
18                 x = x.add(a.divide(x)).divide(two);
19             if (x.mod(two).compareTo(BigInteger.ZERO) == 0) System.out.println(0);
20             else System.out.println(1);
21         }
22         in.close();
23     }
24 }

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C - Travel along the Line

题意：一维坐标系中，刚开始位于原点，有$frac{1}{4}$的概率 坐标 +1 和 -1  有$frac {1}{2} 的概率 不动$  求在第n秒的时候恰好到达第m个位置的概率

思路：考虑把一个0拆成两个0，变成四种操作，这样四种操作是等概率的，那么所有的可能性就是 $4^n$ 再考虑符合条件的方案数

可以考虑将m通过坐标变换转化成正直，那么一个满足题意的操作序列肯定是 1 的个数 减去 -1的 个数 恰好为m

那么我们只需要枚举1的个数，排列组合一下即可

16说 假如用a 表示 1 的个数  b 表示 -1 的个数 c 表示 0的个数

那么有$frac {n!} {a! cdot b! cdot c!}$ 但是这里要考虑 多乘上$2^c$ 因为每个0都有两种选择 ，可以是$0_1 或者 是 0_2$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 100010
 6 ll MOD = (ll)1e9 +7;
 7 
 8 ll fac[N], Bit[N];
 9 ll qmod(ll base, ll n)
10 {
11     ll res = 1;
12     while (n)
13     {
14         if (n & 1) res = res * base % MOD;
15         base = base * base % MOD;
16         n >>= 1;
17     }
18     return res;
19 }
20 
21 void Init()
22 {
23     fac[0] = 1;
24     Bit[0] = 1;
25     for (int i = 1; i < N; ++i) fac[i] = fac[i - 1] * i % MOD;
26     for (int i = 1; i < N; ++i) Bit[i] = Bit[i - 1] * 2 % MOD;
27 }
28 
29 int n, m;
30 
31 int main()
32 {
33     Init();
34     int t; scanf("%d", &t);
35     while (t--)
36     {
37         scanf("%d%d", &n, &m);
38         if (m < 0) m = -m;
39         ll p = 0, q = qmod(4, n); 
40         for (int i = 0; 2 * i + m <= n; ++i)
41             p = (p + (fac[n] * qmod(fac[i], MOD - 2) %MOD * qmod(fac[i + m], MOD - 2) % MOD * qmod(fac[n - 2 * i - m], MOD - 2) % MOD * Bit[n - 2 * i - m] % MOD)) % MOD;    
42         ll res = p * qmod(q, MOD - 2) % MOD;
43         printf("%lld
", res);
44     }
45     return 0;
46 }

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D - Machine Learning on a Tree

留坑。

 

E - Yet Another Tree Query Problem

留坑。

 

F - And Another Data Structure Problem

留坑。

 

G - Neighboring Characters

留坑。

 

H - Happy Sequence ZOJ

题意：用1-n的数，每个数可以用无限次，组成长度为m的序列，求有多少个序列满足 $gcd(b_i, b_{i +1}) = b_{i}$

思路：考虑枚举序列里面不同的数的个数，根据题目范围，最多有10个不同的数，然后隔板法求方案数

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 long long f[15];
 6 long long C[15];
 7 long long MOD;
 8 int p[15];
 9 int n, m;
10 
11 void dp(int t) {
12     int j;
13     j = 2;
14     while (p[t - 1] * j <= n) {
15         p[t] = p[t - 1] * j;
16         f[t]++;
17         dp(t + 1);
18         j++;
19     }
20 }
21 
22 ll qmod(ll base, ll n)
23 {
24     ll res = 1;
25     while (n)
26     {
27         if (n & 1) res = res * base % MOD;
28         base = base * base % MOD;
29         n >>= 1;
30     }
31     return res;
32 }
33 
34 int main()
35 {
36     int i, j, t;
37     long long ans;
38     scanf("%d", &t);
39     MOD = 1000000007;
40     while (t--) {
41         scanf("%d %d", &n, &m);
42         memset(f, 0, sizeof f);
43         memset(p, 0, sizeof p);
44         ans = 0;
45         C[0] = 1;
46         for (i = 1; i <= 11 ; ++i)
47             C[i] = (C[i - 1] * (m - i) % MOD * qmod(i, MOD - 2)) % MOD;
48         for (i = 1; i <= n; ++i) {
49             p[1] = i;
50             ++f[1];
51             dp(2);
52         }
53         for (i = 1; i <= 11; ++i) {
54             ans = (ans + f[i] * C[i - 1] % MOD) % MOD;
55         }
56         printf("%lld
",ans);
57     }
58     return 0;
59 }

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I - Your Bridge is under Attack

 留坑。

 

J - Super Brain

水。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 100010
 5 int n;
 6 int cnt[N * 10], a[N], b[N];
 7 
 8 int main()
 9 {
10     int t; scanf("%d", &t);
11     while (t--)
12     {
13         scanf("%d", &n);
14         for (int i = 1; i <= n; ++i) scanf("%d", a + i);
15         for (int i = 1; i <= n; ++i) scanf("%d", b + i);
16         memset(cnt, 0, sizeof cnt);
17         for (int i = 1; i <= n; ++i) ++cnt[a[i]];
18         int res = 0;
19         for (int i = 1; i <= n; ++i) if (cnt[b[i]] == 1) 
20         {
21             res = b[i];
22             break;
23         }
24         printf("%d
", res);
25     }
26     return 0;
27 }

View Code