A
贪心的取
每个字母n/k次
令r=n%k
让前r个字母各取一次
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 #define rep(i, a, b) for (int i = a; i <= b; ++i)
5
6 int t, n, k;
7
8 int main() {
9 cin >> t;
10
11 while (t--) {
12 cin >> n >> k;
13 int x = n / k, r = n - x * k;
14 rep(i, 1, k) rep(j, 1, x) {
15 cout << (char)(‘a‘ + i - 1);
16 }
17 rep(i, 1, r) {
18 cout << (char)(‘a‘ + i - 1);
19 }
20 cout << ‘\n‘;
21 }
22 return 0;
23 }
B
排序完连续两个比较
证明一下吧：max(a[2] - a[1]), a[4] - a[3]) <= max(a[3] - a[1]), a[4] - a[2]) (后者的间隙大)
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 #define rep(i, a, b) for (int i = a; i <= b; ++i)
5
6 const int N = 1e5 + 7;
7
8 int n, a[N];
9
10 int main() {
11 cin >> n;
12
13 rep(i, 1, n) {
14 cin >> a[i];
15 }
16
17 sort(a + 1, a + n + 1);
18
19 int ans = 0;
20 rep(i, 1, n / 2) {
21 ans += a[i * 2] - a[i * 2 - 1];
22 }
23
24 cout << ans << ‘\n‘;
25
26 return 0;
27 }
C
题意难理解
找出最长的两个串 判断哪个是最长前缀
f[len]代表着长度为len的前后缀是否被选 用来避免长度为len的两个子串都是前缀或都是后缀的情况
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 #define rep(i, a, b) for (ll i = a; i <= b; ++i)
5
6 const int N = 507;
7
8 ll n;
9 bool f[N];
10 string s[N];
11
12 int main() {
13 cin >> n; ll m = 2 * n - 2;
14
15 string s1 = "", s2 = "";
16 rep(i, 1, m) {
17 cin >> s[i];
18 if (s[i].size() > s1.size()) s1 = s[i];
19 else if (s[i].size() == s1.size()) s2 = s[i];
20 }
21
22 ll cnt = 0;
23 rep(i, 1, m) if (s1.substr(0, s[i].size()) == s[i]) {
24 if (s[i] != s2) cnt++;
25 }
26
27 string pre;
28 if (cnt >= n - 1 && s1.substr(1, s1.size() - 1) == s2.substr(0, s1.size() - 1)) pre = s1; else pre = s2;
29
30 rep(i, 1, m) {
31 if (pre.substr(0, s[i].size()) == s[i] && !f[s[i].size()]) {
32 cout << ‘P‘;
33 f[s[i].size()] = 1;
34 }
35 else cout << ‘S‘;
36 }
37
38 return 0;
39 }
D1 D2
两题类似
用一个栈 若当前高度和栈顶一致就弹出 具体判断见代码
1 //D1
2 #include <bits/stdc++.h>
3 using namespace std;
4 typedef long long ll;
5 #define rep(i, a, b) for (int i = a; i <= b; ++i)
6
7 const int N = 2e5 + 7;
8
9 int n;
10 bool a[N];
11 stack <bool> st;
12
13 int main() {
14 scanf("%d", &n);
15
16 int x;
17 rep(i, 1, n) {
18 scanf("%d", &x);
19 a[i] = x & 1;
20 }
21
22 rep(i, 1, n) {
23 if(st.empty())
24 st.push(a[i]);
25 else if(a[i] == st.top())
26 st.pop();
27 else
28 st.push(a[i]);
29 }
30
31 st.size() > 1 ? puts("NO") : puts("YES");
32
33 return 0;
34 }
1 //D2
2 #include <bits/stdc++.h>
3 using namespace std;
4 typedef long long ll;
5 #define rep(i, a, b) for (int i = a; i <= b; ++i)
6
7 int n, mx;
8 stack <int> s;
9
10 int main() {
11 scanf("%d", &n);
12
13 int x;
14 rep(i, 1, n) {
15 scanf("%d",&x);
16 mx=max(mx,x);
17 if (s.empty())
18 s.push(x);
19 else if (s.top()==x)
20 s.pop();
21 else if (s.top()>=x)
22 s.push(x);
23 else {
24 puts("NO");
25 return 0;
26 }
27 }
28
29 if (!s.size())
30 puts("YES");
31 else if (s.size() == 1 && s.top() == mx)
32 puts("YES");
33 else
34 puts("NO");
35
36 return 0;
37 }
E
blank
F
初始思路是：维护树上的一堆和乱搞
其实dfs2的操作十分巧妙
1.res -= sum[v] 代表 v子树下的每个点的距离都少了1
2.sum[u] -= sum[v]; 为下一步做准备
3.res += sum[u]; 代表 u子树下(除)的每个点的距离都多了1
4.sum[v]+=sum[u]; dfs2到下一层的时候 (假如边为（v,w）) res+= sum[v](即步骤3)的时候v点外u点子树下的一些点到w的距离也多了1
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4
5 #define rep(i, a, b) for (int i = a; i <= b; ++i)
6
7 const int N = 2e5 + 7;
8
9 int n;
10
11 ll a[N], sum[N], res, ans;
12
13 vector <int> e[N];
14
15 void dfs1(int u, int p = 0, int d = 0) {
16 res += d * a[u];
17 sum[u] = a[u];
18 for (auto v : e[u]) if (v != p) {
19 dfs1(v, u, d + 1);
20 sum[u] += sum[v];
21 }
22 }
23
24 void dfs2(int u, int p = 0) {
25 ans = max(ans, res);
26 for (auto v : e[u]) if (v != p) {
27 res -= sum[v];
28 sum[u] -= sum[v];
29 res += sum[u];
30 sum[v] += sum[u];
31
32 dfs2(v, u);
33
34 sum[v] -= sum[u];
35 res -= sum[u];
36 sum[u] += sum[v];
37 res += sum[v];
38 }
39 }
40
41 int main() {
42 scanf("%d", &n);
43
44 rep(i, 1, n) {
45 scanf("%lld", &a[i]);
46 }
47
48 int u, v;
49 rep(i, 1, n - 1) {
50 scanf("%d%d", &u, &v);
51 e[u].push_back(v);
52 e[v].push_back(u);
53 }
54
55 dfs1(1);
56 dfs2(1);
57
58 printf("%lld\n", ans);
59
60 return 0;
61 }
原文地址：https://www.cnblogs.com/Fo0o0ol/p/10164365.html
时间： 2024-08-30 14:25:39