傳送門

A

貪心的取

每個字母n/k次

令r=n%k

讓前r個字母各取一次

 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

排序完連續兩個比較

證明一下吧：max(a[2] - a[1]), a[4] - a[3]) <= max(a[3] - a[1]), a[4] - a[2]) (後者的間隙大) 

 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

題意難理解

找出最長的兩個串 判斷哪個是最長字首

f[len]代表著長度為len的前後綴是否被選 用來避免長度為len的兩個子串都是字首或都是字尾的情況

 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

兩題類似

用一個棧 若當前高度和棧頂一致就彈出 具體判斷見程式碼

 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：維護樹上的一堆和亂搞

其實dfs2的操作十分巧妙

1.res -= sum[v] 代表 v子樹下的每個點的距離都少了1

2.sum[u] -= sum[v]; 為下一步做準備

3.res += sum[u]; 代表 u子樹下(除)的每個點的距離都多了1

4.sum[v]+=sum[u]; dfs2到下一層的時候 (假如邊為（v,w）) res+= sum[v](即步驟3)的時候v點外u點子樹下的一些點到w的距離也多了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 }