Educational Codeforces Round 41 (Rated for Div. 2)

这场没打又亏疯了！！！

类似俄罗斯方块，模拟一下就好啦。

 1 #include<bits/stdc++.h>
 2 #define fi first
 3 #define se second
 4 #define mk make_pair
 5 #define pii pair<int,int>
 6 using namespace std;
 7 
 8 typedef long long ll;
 9 const int inf=0x3f3f3f3f;
10 const ll INF=0x3f3f3f3f3f3f3f3f;
11 const int N=1000+7;
12 const int M=12;
13 
14 int n, m;
15 
16 int cnt[N];
17 int main() {
18 
19     int ans = 0;
20     scanf("%d%d", &n, &m);
21     for(int i = 1; i <= m; i++) {
22         int x; scanf("%d", &x);
23         cnt[x]++;
24 
25         bool flag = true;
26         for(int j = 1; j <= n; j++) {
27             if(cnt[j] == 0)
28                 flag = false;
29         }
30 
31         if(flag) {
32             ans++;
33             for(int j = 1; j <= n; j++) {
34                 cnt[j] -= 1;
35             }
36         }
37     }
38 
39     printf("%d
", ans);
40     return 0;
41 }
42 /*
43 */

View Code

B - Lecture Sleep：

瞎模拟一下就好啦。

 1 #include<bits/stdc++.h>
 2 #define fi first
 3 #define se second
 4 #define mk make_pair
 5 #define pii pair<int,int>
 6 using namespace std;
 7 
 8 typedef long long ll;
 9 const int inf=0x3f3f3f3f;
10 const ll INF=0x3f3f3f3f3f3f3f3f;
11 const int N=1e5+7;
12 const int M=12;
13 
14 int n, k, a[N], t[N],sum1[N], sum2[N];
15 
16 int main() {
17 
18     scanf("%d%d", &n, &k);
19 
20     for(int i = 1; i <= n; i++) {
21         scanf("%d", &a[i]);
22         sum1[i] = sum1[i-1] + a[i];
23     }
24 
25     for(int i = 1; i <= n; i++) {
26         scanf("%d", &t[i]);
27         sum2[i] = sum2[i-1];
28         if(t[i])
29             sum2[i] += a[i];
30     }
31 
32     int ans = sum2[n];
33 
34     for(int i = 1; i + k -1 <= n; i++) {
35         ans = max(ans, sum2[n] + (sum1[i+k-1] - sum1[i-1]) - (sum2[i+k-1] - sum2[i-1]));
36     }
37 
38     printf("%d
",ans);
39     return 0;
40 }
41 /*
42 */

View Code

C - Chessboard：

每个块只有两种情况，第一种是第一个数字为0，第二种是第一个数字为1，把每个块这两种情况需要修改的个数记录下来，

然后暴力枚举就好啦。

 1 #include<bits/stdc++.h>
 2 #define fi first
 3 #define se second
 4 #define mk make_pair
 5 #define pii pair<int,int>
 6 using namespace std;
 7 
 8 typedef long long ll;
 9 const int inf=0x3f3f3f3f;
10 const ll INF=0x3f3f3f3f3f3f3f3f;
11 const int N=1e5+7;
12 const int M=12;
13 
14 int n, a[4][2], s[101][101],id[4];
15 int main() {
16 
17     scanf("%d", &n);
18 
19     for(int k = 0; k < 4; k++) {
20         for(int i = 0; i < n; i++) {
21             for(int j = 0; j < n; j++) {
22                 scanf("%1d", &s[i][j]);
23             }
24         }
25 
26         int flag = 0;
27         for(int i = 0; i < n; i++) {
28             for(int j = 0; j < n; j++) {
29                 if(flag != s[i][j])
30                     a[k][0]++;
31                 flag ^= 1;
32             }
33         }
34 
35         flag = 1;
36         for(int i = 0; i < n; i++) {
37             for(int j = 0; j < n; j++) {
38                 if(flag !=s[i][j])
39                     a[k][1]++;
40                 flag ^= 1;
41             }
42         }
43     }
44     for(int i = 0; i < 4; i++)
45         id[i] = i;
46 
47     int ans = inf;
48     do
49     {
50         ans = min(ans, a[id[0]][0] + a[id[1]][1] + a[id[2]][0] + a[id[3]][1]);
51     }while(next_permutation(id, id + 4));
52 
53     for(int i = 0; i < 4; i++)
54         id[i] = i;
55 
56     do
57     {
58         ans = min(ans, a[id[0]][1] + a[id[1]][0] + a[id[2]][1] + a[id[3]][0]);
59     }while(next_permutation(id, id + 4));
60 
61     printf("%d
", ans);
62     return 0;
63 }
64 /*
65 */

View Code

D - Pair Of Lines：

题意：给你若干个点，问你能不能最多划两条直线覆盖所有点。

思路：先找到三个不共线的点组成一个三角形，其中一条边一定是需要划的直线，枚举一下这三条边，然后check一下剩下的

点在不在一条直线上。

  1 #include<bits/stdc++.h>
  2 #define fi first
  3 #define se second
  4 #define mk make_pair
  5 #define pii pair<int,int>
  6 using namespace std;
  7 
  8 typedef long long LL;
  9 const int inf=0x3f3f3f3f;
 10 const LL INF=0x3f3f3f3f3f3f3f3f;
 11 const int N=1e5+7;
 12 const int M=12;
 13 
 14 struct point {
 15     LL x, y;
 16 }p[N];
 17 
 18 LL aross(point a, point b, point c) {
 19     return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
 20 }
 21 
 22 vector<point> v;
 23 int n;
 24 int main() {
 25 
 26     scanf("%d", &n);
 27 
 28     if(n <= 4) {
 29         puts("YES");
 30         return 0;
 31     }
 32     for(int i = 0; i < n; i++) {
 33         scanf("%lld", &p[i].x);
 34         scanf("%lld", &p[i].y);
 35     }
 36 
 37     int pos = 2;
 38 
 39     while(pos < n && !aross(p[0], p[1], p[pos]))
 40         pos++;
 41 
 42     if(pos == n) {
 43         puts("YES");
 44         return 0;
 45     } else {
 46 
 47         for(int i = 0; i < n; i++) {
 48             if(aross(p[0], p[1], p[i]) != 0)
 49                 v.push_back(p[i]);
 50         }
 51 
 52         if(v.size() <= 2) {
 53             puts("YES");
 54             return 0;
 55         }
 56 
 57         bool flag = true;
 58 
 59         for(int i = 2; i < v.size(); i++) {
 60             if(aross(v[0], v[1], v[i]) !=0) {
 61                 flag = false;
 62                 break;
 63             }
 64         }
 65 
 66         if(flag) {
 67             puts("YES");
 68             return 0;
 69         }
 70 
 71         v.clear();
 72         for(int i = 0; i < n; i++) {
 73             if(aross(p[0], p[pos], p[i]) != 0)
 74                 v.push_back(p[i]);
 75         }
 76 
 77         if(v.size() <= 2) {
 78             puts("YES");
 79             return 0;
 80         }
 81 
 82         flag = true;
 83 
 84         for(int i = 2; i < v.size(); i++) {
 85             if(aross(v[0], v[1], v[i]) !=0) {
 86                 flag = false;
 87                 break;
 88             }
 89         }
 90 
 91         if(flag) {
 92             puts("YES");
 93             return 0;
 94         }
 95 
 96         
 97         v.clear();
 98         for(int i = 0; i < n; i++) {
 99             if(aross(p[1], p[pos], p[i]) != 0)
100                 v.push_back(p[i]);
101         }
102 
103         if(v.size() <= 2) {
104             puts("YES");
105             return 0;
106         }
107 
108         flag = true;
109 
110         for(int i = 2; i < v.size(); i++) {
111             if(aross(v[0], v[1], v[i]) !=0) {
112                 flag = false;
113                 break;
114             }
115         }
116 
117         if(flag) {
118             puts("YES");
119             return 0;
120         }
121 
122         puts("NO");
123     }
124     return 0;
125 }
126 /*
127 */

View Code

E - Tufurama：

裸的主席树，没啥好说的，数组开小了两次一次WA，一次RE。。。

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 #define fi first
 4 #define se second
 5 #define mk make_pair
 6 #define pii pair<int,int>
 7 using namespace std;
 8 
 9 typedef long long LL;
10 const int inf=0x3f3f3f3f;
11 const LL INF=0x3f3f3f3f3f3f3f3f;
12 const int N=4e5+7;
13 const int M=12;
14 
15 int root[N],hs[N],a[N],tot;
16 struct Chairman_tree
17 {
18     int cnt;
19     struct node{
20         int l,r;
21         ll sum;
22     }a[20*N];
23     void update(int l,int r,int &x,int y,int pos,int v)
24     {
25         a[++cnt]=a[y];
26         x=cnt; a[x].sum+=v;
27         if(l==r) return;
28         int mid=(l+r)>>1;
29         if(pos<=mid)
30             update(l,mid,a[x].l,a[y].l,pos,v);
31         else
32             update(mid+1,r,a[x].r,a[y].r,pos,v);
33     }
34     ll query(int l,int r,int L,int R,int x)
35     {
36         if(l >= L && r <= R)
37             return a[x].sum;
38         ll ans = 0;
39         int mid = (l + r) >> 1;
40         if(L <= mid)
41             ans += query(l,mid,L,R,a[x].l);
42         if(R > mid)
43             ans += query(mid + 1,r,L,R,a[x].r);
44         return ans;
45     }
46 }seg;
47 
48 int n;
49 int main() {
50 
51     scanf("%d", &n);
52 
53     for(int i = 1; i <= n; i++) {
54         scanf("%d", &a[i]);
55         hs[++tot] = a[i];
56         hs[++tot] = i;
57     }
58 
59     sort(hs + 1, hs + tot + 1);
60     tot = unique(hs + 1, hs + tot + 1) - hs - 1;
61 
62     for(int i = 1; i <= n; i++) {
63         int pos = lower_bound(hs+1,hs+1+tot,a[i])-hs;
64         seg.update(1,tot,root[i],root[i-1],pos,1);
65     }
66 
67     ll ans = 0;
68     for(int i = 1; i <= n; i++) {
69         int item = lower_bound(hs+1,hs+1+tot,i)-hs;
70         int pos = min(i - 1, a[i]);
71         ans += seg.query(1, tot, item, tot, root[pos]);
72     }
73 
74     printf("%lld
",ans);
75     return 0;
76 }
77 /*
78 */

View Code

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原文地址：https://www.cnblogs.com/CJLHY/p/8724053.html