Codeforces Edu Round 68 (Rated for Div. 2)

比较水的一场。

题目链接：https://codeforces.com/contest/1194

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A：

秒懂跟x没关系，答案就是2n。

B：

因为n*m<=4e5，统计每行每列点的数量后直接O(nm)暴力枚举维护最小值即可。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int main() {
21     int t; cin >> t;
22     while (t--) {
23         int n, m, ans = int_inf; cin >> n >> m;
24         char s[n + 1][m + 1];
25         int row[n + 1] = {0}, col[m + 1] = {0};
26         rep1(i, 1, n) {
27             scanf("%s", s[i] + 1);
28             rep1(j, 1, m) {
29                 if (s[i][j] == '.') row[i]++, col[j]++;
30             }
31         }
32         rep1(i, 1, n) {
33             rep1(j, 1, m) {
34                 int k = row[i] + col[j];
35                 if (s[i][j] == '.') k--;
36                 ans = min(ans, k);
37             }
38         }
39         printf("%d\n", ans);
40     }
41     return 0;
42 }

View Code

C：

O(n)扫一遍就行了。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 110;
21 int q;
22 
23 int main() {
24     scanf("%d", &q);
25     while (q--) {
26         string s, t, p; cin >> s >> t >> p;
27         int len1 = s.size(), len2 = t.size(), len3 = p.size(), ans = 1;
28         int a[26] = {0};
29         for (auto c : p) a[c - 'a']++;
30         int i = 0, j = 0;
31         for (; i < len2; i++) {
32             if (j < len1 && t[i] == s[j]) j++;
33             else if (a[t[i] - 'a']) a[t[i] - 'a']--;
34             else {
35                 ans = 0;
36                 break;
37             }
38         }
39         if (j != len1) ans = 0;
40         if (ans) puts("YES"); else puts("NO");
41     }
42     return 0;
43 }

View Code

D：

一看就是SG函数相关的题，打个表找规律发现循环节长度为3。简单推一下必胜必败点也可以猜出来。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int t;
21 
22 int main() {
23     scanf("%d", &t);
24     while (t--) {
25         int n, k; scanf("%d%d", &n, &k);
26         if (k % 3 == 0) {
27             int p = n % (k + 1) + 1;
28             if (p == k + 1) {
29                 puts("Alice");
30                 continue;
31             } else {
32                 if (p % 3 == 1) puts("Bob");
33                 else puts("Alice");
34             }
35         } else {
36             if (n % 3 == 0) puts("Bob");
37             else puts("Alice");
38         }
39     }
40     return 0;
41 }

View Code

E：

树状数组+扫描线。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 1e4 + 10;
21 int n, m, tot1 = 0, tot2 = 0, c[maxn], vis[maxn];
22 ll ans = 0;
23 
24 struct Vert {
25     int x, y, xx;
26     bool operator<(const Vert &rhs)const {
27         return y < rhs.y;
28     }
29 } v[maxn];
30 
31 struct Hori {
32     int x, y, yy;
33     bool operator<(const Hori &rhs)const {
34         return yy < rhs.yy;
35     }
36 } h[maxn];
37 
38 int lowbit(int x) {
39     return x & -x;
40 }
41 
42 int find(int x) {
43     int ret = 0;
44     while (x > 0) {
45         ret += c[x];
46         x -= lowbit(x);
47     }
48     return ret;
49 }
50 
51 void add(int x, int val) {
52     while (x <= 10001) {
53         c[x] += val;
54         x += lowbit(x);
55     }
56 }
57 
58 int main() {
59     scanf("%d", &n);
60     rep1(i, 1, n) {
61         int x, y, xx, yy; scanf("%d%d%d%d", &x, &y, &xx, &yy);
62         x += 5001, y += 5001, xx += 5001, yy += 5001;
63         if (x == xx) h[++tot2] = (Hori) {
64             x, min(y, yy), max(y, yy)
65         };
66         else v[++tot1] = (Vert) {
67             min(x, xx), y, max(x, xx)
68         };
69     }
70     sot(v, tot1); sot(h, tot2);
71     sort(v + 1, v + 1 + tot1);
72     sort(h + 1, h + 1 + tot2);
73     rep1(i, 1, tot1) {
74         rep0(j, 0, maxn) c[j] = 0;
75         rep1(j, 1, tot2) {
76             if (h[j].y <= v[i].y && h[j].yy >= v[i].y) {
77                 add(h[j].x, 1);
78                 vis[j] = 1;
79             } else vis[j] = 0;
80         }
81         int k = 1;
82         rep1(j, i + 1, tot1) {
83             while (k <= tot2 && h[k].yy < v[j].y) {
84                 if (vis[k]) add(h[k].x, -1);
85                 k++;
86             }
87             int l = max(v[i].x, v[j].x) - 1, r = min(v[i].xx, v[j].xx);
88             if (l < r) {
89                 int x = find(r) - find(l);
90                 ans += 1LL * x * (x - 1) / 2;
91             }
92         }
93     }
94     printf("%lld\n", ans);
95     return 0;
96 }

View Code

F && G：

不会，溜了(

 

转载于:https://www.cnblogs.com/JHSeng/p/11190004.html