Ivan_Chien
C++
A
签到题。
1#include <iostream>
2#include <climits>
3
4int main()
5{
6 int n, t;
7 std::cin >> n >> t;
8
9 int prev = INT_MIN / 2;
10 int curr;
11 while (std::cin >> curr)
12 {
13 if (curr - prev <= t)
14 {
15 std::cout << curr << std::endl;
16 return 0;
17 }
18 prev = curr;
19 }
20
21 std::cout << -1 << std::endl;
22
23 return 0;
24}
B
模拟。有多对可消时，易证消的顺序不影响最后结果。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n;
7 std::cin >> n;
8 std::vector<int> v(n);
9
10 for (int i = 0; i < n; ++i)
11 {
12 std::cin >> v[i];
13 }
14
15 i64 ans = 0;
16 bool flag = true;
17 while (flag)
18 {
19 flag = false;
20 for (int i = 0; i + 1 < n; ++i)
21 {
22 if (v[i] == v[i + 1])
23 {
24 flag = true;
25 ans += 1LL * v[i];
26 v.erase(v.begin() + i);
27 v.erase(v.begin() + i);
28 n -= 2;
29 break;
30 }
31 }
32 }
33
34 std::cout << ans << std::endl;
35
36 return 0;
37}
C
两两相消，使用 map 来维护数的库存。排序后用双指针亦可。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n, sum;
7 std::cin >> n >> sum;
8
9 std::map<int, int> cnt;
10 i64 ans = 0;
11 for (int i = 0; i < n; ++i)
12 {
13 int a;
14 std::cin >> a;
15 if (cnt[sum - a] > 0)
16 {
17 --cnt[sum - a];
18 ++ans;
19 }
20 else
21 {
22 ++cnt[a];
23 }
24 }
25
26 std::cout << ans << std::endl;
27
28 return 0;
29}
D
统计字母数量是否足够即可。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n, m;
7 std::string a, b;
8 std::cin >> n >> m >> a >> b;
9
10 std::map<char, int> cnt;
11 for (auto c : a)
12 {
13 ++cnt[c];
14 }
15
16 for (auto c : b)
17 {
18 if (cnt[c] <= 0)
19 {
20 std::cout << "NO" << std::endl;
21 return 0;
22 }
23 --cnt[c];
24 }
25
26 std::cout << "YES" << std::endl;
27
28 return 0;
29}
E
贪心。从 0 出发，对石头高度排序，每次往序列的另一头跳即可。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n, m;
7 std::cin >> n >> m;
8 std::vector<int> v(n);
9
10 for (int i = 0; i < n; ++i)
11 {
12 std::cin >> v[i];
13 }
14
15 std::sort(v.begin(), v.end());
16
17 int left = 0, right = n;
18 int currH = 0;
19 i64 ans = 0;
20 while (left < right)
21 {
22 int d1 = std::abs(currH - v[left]);
23 int d2 = std::abs(currH - v[right - 1]);
24 if (d1 > d2)
25 {
26 ans += 1LL * d1;
27 currH = v[left];
28 ++left;
29 }
30 else // d1 < d2, NEVER EQUAL
31 {
32 ans += 1LL * d2;
33 currH = v[right - 1];
34 --right;
35 }
36 }
37
38 if (ans <= m)
39 {
40 std::cout << "算你厉害" << std::endl;
41 }
42 else
43 {
44 std::cout << ans - m << std::endl;
45 }
46
47 return 0;
48}
F
枚举所有区间和即可，用前缀和做个简单优化。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n, m;
7 std::cin >> n >> m;
8
9 std::vector<int> pref(n + 1);
10 for (int i = 1; i <= n; ++i)
11 {
12 int a;
13 std::cin >> a;
14 pref[i] = a + pref[i - 1];
15 }
16
17 i64 ans = 0;
18 for (int i = 1; i <= n; ++i)
19 {
20 for (int j = 0; j < i; ++j)
21 {
22 if (pref[i] - pref[j] == m)
23 {
24 ++ans;
25 }
26 }
27 }
28
29 std::cout << ans << std::endl;
30
31 return 0;
32}
G
模拟。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 int n, m;
7 std::cin >> n >> m;
8
9 std::vector mat(n, std::vector(m, 0));
10
11 for (auto& v : mat)
12 {
13 for (auto& a : v)
14 {
15 std::cin >> a;
16 }
17 }
18
19 int cq;
20 std::cin >> cq;
21 std::vector<int> q(m);
22 while (cq--)
23 {
24 int ans = 0;
25 for (int i = 0; i < m; ++i)
26 {
27 std::cin >> q[i];
28 }
29
30 for (int i = 0; i < n; ++i)
31 {
32 bool flag = true;
33 for (int j = 0; j < m; ++j)
34 {
35 if (q[j] != -1 && q[j] != mat[i][j])
36 {
37 flag = false;
38 break;
39 }
40 }
41 if (flag) ++ans;
42 }
43 std::cout << ans << std::endl;
44 }
45
46 return 0;
47}
H
简单构造。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 i64 n, m;
7 std::cin >> n >> m;
8 i64 ans = 0;
9
10 while (n != m)
11 {
12 if (n < m)
13 {
14 std::swap(n, m);
15 }
16 i64 f = n / m;
17 if (f * m == n) --f;
18 n -= f * m;
19 ans += f;
20 }
21
22 std::cout << ans << std::endl;
23
24 return 0;
25}
i
前缀和。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 i64 n, w;
7 std::cin >> n >> w;
8
9 std::vector<int> pref(n + 1);
10 for (int i = 1; i <= n; ++i)
11 {
12 int a;
13 std::cin >> a;
14 pref[i] = a + pref[i - 1];
15 }
16
17 int max = -1;
18 for (int i = w; i <= n; ++i)
19 {
20 max = std::max(max, pref[i] - pref[i - w]);
21 }
22 std::cout << max << std::endl;
23
24 return 0;
25}
J
数据范围小，直接暴力。正统做法大概是 unique 后做区间 DP。
1#include <bits/stdc++.h>
2using i64 = long long;
3
4int main()
5{
6 std::string s;
7 std::cin >> s;
8
9 int ans_odd = 0, ans_even = 0;
10 int n = s.length();
11 for (int d = 1; d <= n; ++d)
12 {
13 for (int i = 0; i + d <= n; ++i)
14 {
15 auto ss = s.substr(i, d);
16 ss.erase(std::unique(ss.begin(), ss.end()), ss.end());
17 int nn = ss.length();
18 bool flag = true;
19 for (int i = 0; i < nn / 2; ++i)
20 {
21 if (ss[i] != ss[nn - 1 - i])
22 {
23 flag = false;
24 break;
25 }
26 }
27 if (flag)
28 {
29 if (d & 1) ++ans_odd;
30 else ++ans_even;
31 }
32 }
33 }
34
35 std::cout << ans_even << ' ' << ans_odd << std::endl;
36
37 return 0;
38}
Python 参考
C
1from collections import Counter
2
3n, m = map(int, input().split())
4c = Counter()
5ans = 0
6for _ in range(n):
7 a = int(input())
8 if c[m - a] > 0:
9 c[m - a] -= 1
10 ans += 1
11 else:
12 c[a] += 1
13print(ans)
Java 参考
A
1import java.util.Scanner;
2
3public class Main {
4 public static void main(String[] args) {
5 Scanner s = new Scanner(System.in);
6 int n = s.nextInt();
7 int m = s.nextInt();
8 int prev = -114514191;
9 for (int i = 0; i < n; i++) {
10 int a;
11 a = s.nextInt();
12 if (a - prev <= m) {
13 System.out.println(a);
14 return ;
15 }
16 prev = a;
17 }
18 System.out.println(-1);
19 }
20}