Codeforces Round #590 (Div. 3)

传送门

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 105;
 5 double a[maxn];
 6 int main() {
 7     int q; scanf("%d",&q);
 8     while (q--) {
 9         int n; scanf("%d",&n);
10         for (int i = 1; i <= n; ++i) {
11             scanf("%lf",&a[i]);
12         }
13         double sum = 0;
14         for (int i = 1; i <= n; ++i) sum += a[i];
15         int ans = ceil(sum / n);
16         printf("%d\n",ans);
17     }
18     return 0;
19 }
A. Equalize Prices Again

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5+5;
 5 bool vis[maxn];
 6 int a[maxn], b[maxn];
 7 deque<int> que;
 8 int main() {
 9     int n, k; scanf("%d%d",&n,&k);
10     for (int i = 1; i <= n; ++i) {
11         scanf("%d",&a[i]);
12         b[i] = a[i];
13     }
14     sort(b+1,b+1+n);
15     int m = unique(b+1,b+1+n) - b - 1;
16     for (int i = 1; i <= n; ++i) {
17         int id = lower_bound(b+1,b+1+m,a[i]) - b;
18         if (vis[id] == false) {
19             vis[id] = true;
20             if (que.size() < k) que.push_front(id);
21             else {
22                 vis[que.back()] = false;
23                 que.pop_back();
24                 que.push_front(id);
25             }
26         }
27         else continue;
28     }
29     printf("%d\n",que.size());
30     while (!que.empty()) {
31         int id = que.front(); que.pop_front();
32         printf("%d ",b[id]);
33     }
34     printf("\n");
35     return 0;
36 }
B1. Social Network (easy version)

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5+5;
 5 bool vis[maxn];
 6 int a[maxn], b[maxn];
 7 deque<int> que;
 8 int main() {
 9     int n, k; scanf("%d%d",&n,&k);
10     for (int i = 1; i <= n; ++i) {
11         scanf("%d",&a[i]);
12         b[i] = a[i];
13     }
14     sort(b+1,b+1+n);
15     int m = unique(b+1,b+1+n) - b - 1;
16     for (int i = 1; i <= n; ++i) {
17         int id = lower_bound(b+1,b+1+m,a[i]) - b;
18         if (vis[id] == false) {
19             vis[id] = true;
20             if (que.size() < k) que.push_front(id);
21             else {
22                 vis[que.back()] = false;
23                 que.pop_back();
24                 que.push_front(id);
25             }
26         }
27         else continue;
28     }
29     printf("%d\n",que.size());
30     while (!que.empty()) {
31         int id = que.front(); que.pop_front();
32         printf("%d ",b[id]);
33     }
34     printf("\n");
35     return 0;
36 }
B2. Social Network (hard version)

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5+5;
 5 string s1, s2;
 6 int main() {
 7     int q; scanf("%d",&q);
 8     while (q--) {
 9         int n; scanf("%d",&n);
10         cin >> s1 >> s2;
11         int now_x = 0, now_y = 0;
12         bool flag = true;
13         while (true) {
14             if (now_y == n) {
15                 if (now_x == 0) flag = false;
16                 break;
17             }
18             if (now_x == 0) {
19                 if (s1[now_y] <= '2') {
20                     now_y++;
21                 }
22                 else {
23                     if (s2[now_y] <= '2') {
24                         flag = false;
25                         break;
26                     }
27                     now_x = 1, now_y++;
28                 }
29             }
30             else {
31                 if (s2[now_y] <= '2') {
32                     now_y++;
33                 }
34                 else {
35                     if (s1[now_y] <= '2') {
36                         flag = false;
37                         break;
38                     }
39                     now_x = 0, now_y++;
40                 }
41             }
42         }
43         if (flag) puts("YES");
44         else puts("NO");
45     }
46     return 0;
47 }
C. Pipes

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 int C[30][100005];
 5 char s[100005];
 6 int lowbit(int x) {
 7     return x&(-x);
 8 }
 9 void add(int ch, int x, int d) {
10     while (x <= 100000) {
11         C[ch][x] += d;
12         x += lowbit(x);
13     }
14 }
15 int sum(int ch, int x) {
16     int res = 0;
17     while (x > 0) {
18         res += C[ch][x];
19         x -= lowbit(x);
20     }
21     return res;
22 }
23 int main() {
24     scanf("%s",s+1);
25     int len = strlen(s+1);
26     for (int i = 1; i <= len; ++i) {
27         add(s[i]-'a',i,1);
28     }
29     int q; scanf("%d",&q);
30     while (q--) {
31         int op; scanf("%d",&op);
32         if (op == 1) {
33             int pos; char ch;
34             scanf("%d",&pos); getchar();
35             ch = getchar();
36             add(s[pos]-'a',pos,-1);
37             s[pos] = ch;
38             add(s[pos]-'a',pos,1);
39         }
40         else {
41             int l, r; scanf("%d%d",&l,&r);
42             int ans = 0;
43             for (char ch = 'a'; ch <= 'z'; ++ch) {
44                 if (sum(ch-'a',r)-sum(ch-'a',l-1) > 0) ans++;
45             }
46             printf("%d\n",ans);
47         }
48     }
49     return 0;
50 }
D. Distinct Characters Queries

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5+5;
 5 int a[maxn], p[maxn];
 6 vector<int> ve[maxn];
 7 int main() {
 8     int n, m; scanf("%d%d",&n,&m);
 9     for (int i = 1; i <= n; ++i) {
10         p[i] = i;
11     }
12     for (int i = 1; i <= m; ++i) {
13         scanf("%d",&a[i]);
14         ve[a[i]].push_back(i);
15     }
16     ll ans = 0;
17     for (int i = 1; i < m; ++i) {
18         ans += abs(a[i]-a[i+1]);
19     }
20     printf("%lld",ans);
21 
22     for (int i = 2; i <= n; ++i) {
23         for (int j = 0; j < ve[i].size(); ++j) {
24             int pos = ve[i][j];
25             if (pos > 1) ans -= abs(i - p[a[pos-1]]);
26             if (pos < m) ans -= abs(i - p[a[pos+1]]);
27         }
28         for (int j = 0; j < ve[i-1].size(); ++j) {
29             int pos = ve[i-1][j];
30             if (pos > 1 && a[pos-1] != i) ans -= abs(1 - p[a[pos-1]]);
31             if (pos < m && a[pos+1] != i) ans -= abs(1 - p[a[pos+1]]);
32         }
33         p[i] = 1; p[i-1] = i;
34         for (int j = 0; j < ve[i].size(); ++j) {
35             int pos = ve[i][j];
36             if (pos > 1) ans += abs(1 - p[a[pos-1]]);
37             if (pos < m) ans += abs(1 - p[a[pos+1]]);
38         }
39         for (int j = 0; j < ve[i-1].size(); ++j) {
40             int pos = ve[i-1][j];
41             if (pos > 1 && a[pos-1] != i) ans += abs(i - p[a[pos-1]]);
42             if (pos < m && a[pos+1] != i) ans += abs(i - p[a[pos+1]]);
43         }
44         printf(" %lld",ans);
45     }
46     printf("\n");
47     return 0;
48 }
E. Special Permutations

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e6+5;
 5 const int _up = (1<<20);
 6 char s[maxn];
 7 int dp[maxn];
 8 int main() {
 9     scanf("%s",s+1);
10     int len = strlen(s+1);
11     for (int i = 1; i <= len; ++i) {
12         int now = 0;
13         for (int j = i; j <= len; ++j) {
14             if (now & (1 << (s[j]-'a'))) break;
15             now |= (1 << (s[j]-'a'));
16             dp[now] = j - i + 1;
17         }
18     }
19     for (int i = 1; i < _up; ++i) {
20         for (int j = 0; j < 20; ++j) {
21             int tmp = (1 << j);
22             if (i & tmp) {
23                 dp[i] = max(dp[i], dp[i^tmp]);
24             }
25         }
26     }
27     int ans = 0;
28     for (int i = 1; i < _up; ++i) {
29         ans = max(ans, dp[i]+dp[_up - 1 - i]);
30     }
31     printf("%d\n",ans);
32     return 0;
33 }
F. Yet Another Substring Reverse

 

posted @ 2019-11-24 13:54  麻辣猪仔  阅读(113)  评论(0)  编辑  收藏  举报