Educational Codeforces Round 41 (Rated for Div. 2)

这场没打又亏疯了！！！

A - Tetris ：

类似俄罗斯方块，模拟一下就好啦。

 #include<bits/stdc++.h>
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 using namespace std;

 typedef long long ll;
 const int inf=0x3f3f3f3f;
 const ll INF=0x3f3f3f3f3f3f3f3f;
 const int N=+;
 const int M=;

 int n, m;

 int cnt[N];
 int main() {

     int ans = ;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= m; i++) {
         int x; scanf("%d", &x);
         cnt[x]++;

         bool flag = true;
         for(int j = ; j <= n; j++) {
             if(cnt[j] == )
                 flag = false;
         }

         if(flag) {
             ans++;
             for(int j = ; j <= n; j++) {
                 cnt[j] -= ;
             }
         }
     }

     printf("%d\n", ans);
     return ;
 }
 /*
 */

B - Lecture Sleep：

瞎模拟一下就好啦。

 #include<bits/stdc++.h>
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 using namespace std;

 typedef long long ll;
 const int inf=0x3f3f3f3f;
 const ll INF=0x3f3f3f3f3f3f3f3f;
 const int N=1e5+;
 const int M=;

 int n, k, a[N], t[N],sum1[N], sum2[N];

 int main() {

     scanf("%d%d", &n, &k);

     for(int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         sum1[i] = sum1[i-] + a[i];
     }

     for(int i = ; i <= n; i++) {
         scanf("%d", &t[i]);
         sum2[i] = sum2[i-];
         if(t[i])
             sum2[i] += a[i];
     }

     int ans = sum2[n];

     for(int i = ; i + k - <= n; i++) {
         ans = max(ans, sum2[n] + (sum1[i+k-] - sum1[i-]) - (sum2[i+k-] - sum2[i-]));
     }

     printf("%d\n",ans);
     return ;
 }
 /*
 */

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 。

 #include<bits/stdc++.h>
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 using namespace std;

 typedef long long ll;
 const int inf=0x3f3f3f3f;
 const ll INF=0x3f3f3f3f3f3f3f3f;
 const int N=1e5+;
 const int M=;

 int n, a[][], s[][],id[];
 int main() {

     scanf("%d", &n);

     for(int k = ; k < ; k++) {
         for(int i = ; i < n; i++) {
             for(int j = ; j < n; j++) {
                 scanf("%1d", &s[i][j]);
             }
         }

         int flag = ;
         for(int i = ; i < n; i++) {
             for(int j = ; j < n; j++) {
                 if(flag != s[i][j])
                     a[k][]++;
                 flag ^= ;
             }
         }

         flag = ;
         for(int i = ; i < n; i++) {
             for(int j = ; j < n; j++) {
                 if(flag !=s[i][j])
                     a[k][]++;
                 flag ^= ;
             }
         }
     }
     for(int i = ; i < ; i++)
         id[i] = i;

     int ans = inf;
     do
     {
         ans = min(ans, a[id[]][] + a[id[]][] + a[id[]][] + a[id[]][]);
     }while(next_permutation(id, id + ));

     for(int i = ; i < ; i++)
         id[i] = i;

     do
     {
         ans = min(ans, a[id[]][] + a[id[]][] + a[id[]][] + a[id[]][]);
     }while(next_permutation(id, id + ));

     printf("%d\n", ans);
     return ;
 }
 /*
 */

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。

 #include<bits/stdc++.h>
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 using namespace std;

 typedef long long LL;
 const int inf=0x3f3f3f3f;
 const LL INF=0x3f3f3f3f3f3f3f3f;
 const int N=1e5+;
 const int M=;

 struct point {
     LL x, y;
 }p[N];

 LL aross(point a, point b, point c) {
     return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
 }

 vector<point> v;
 int n;
 int main() {

     scanf("%d", &n);

     if(n <= ) {
         puts("YES");
         return ;
     }
     for(int i = ; i < n; i++) {
         scanf("%lld", &p[i].x);
         scanf("%lld", &p[i].y);
     }

     int pos = ;

     while(pos < n && !aross(p[], p[], p[pos]))
         pos++;

     if(pos == n) {
         puts("YES");
         return ;
     } else {

         for(int i = ; i < n; i++) {
             if(aross(p[], p[], p[i]) != )
                 v.push_back(p[i]);
         }

         if(v.size() <= ) {
             puts("YES");
             return ;
         }

         bool flag = true;

         for(int i = ; i < v.size(); i++) {
             if(aross(v[], v[], v[i]) !=) {
                 flag = false;
                 break;
             }
         }

         if(flag) {
             puts("YES");
             return ;
         }

         v.clear();
         for(int i = ; i < n; i++) {
             if(aross(p[], p[pos], p[i]) != )
                 v.push_back(p[i]);
         }

         if(v.size() <= ) {
             puts("YES");
             return ;
         }

         flag = true;

         for(int i = ; i < v.size(); i++) {
             if(aross(v[], v[], v[i]) !=) {
                 flag = false;
                 break;
             }
         }

         if(flag) {
             puts("YES");
             return ;
         }

         v.clear();
         for(int i = ; i < n; i++) {
             if(aross(p[], p[pos], p[i]) != )
                 v.push_back(p[i]);
         }

         if(v.size() <= ) {
             puts("YES");
             return ;
         }

         flag = true;

         for(int i = ; i < v.size(); i++) {
             if(aross(v[], v[], v[i]) !=) {
                 flag = false;
                 break;
             }
         }

         if(flag) {
             puts("YES");
             return ;
         }

         puts("NO");
     }
     return ;
 }
 /*
 */

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。。。

 #include<bits/stdc++.h>
 #define ll long long
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 using namespace std;

 typedef long long LL;
 const int inf=0x3f3f3f3f;
 const LL INF=0x3f3f3f3f3f3f3f3f;
 const int N=4e5+;
 const int M=;

 int root[N],hs[N],a[N],tot;
 struct Chairman_tree
 {
     int cnt;
     struct node{
         int l,r;
         ll sum;
     }a[*N];
     void update(int l,int r,int &x,int y,int pos,int v)
     {
         a[++cnt]=a[y];
         x=cnt; a[x].sum+=v;
         if(l==r) return;
         int mid=(l+r)>>;
         if(pos<=mid)
             update(l,mid,a[x].l,a[y].l,pos,v);
         else
             update(mid+,r,a[x].r,a[y].r,pos,v);
     }
     ll query(int l,int r,int L,int R,int x)
     {
         if(l >= L && r <= R)
             return a[x].sum;
         ll ans = ;
         int mid = (l + r) >> ;
         if(L <= mid)
             ans += query(l,mid,L,R,a[x].l);
         if(R > mid)
             ans += query(mid + ,r,L,R,a[x].r);
         return ans;
     }
 }seg;

 int n;
 int main() {

     scanf("%d", &n);

     for(int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         hs[++tot] = a[i];
         hs[++tot] = i;
     }

     sort(hs + , hs + tot + );
     tot = unique(hs + , hs + tot + ) - hs - ;

     for(int i = ; i <= n; i++) {
         int pos = lower_bound(hs+,hs++tot,a[i])-hs;
         seg.update(,tot,root[i],root[i-],pos,);
     }

     ll ans = ;
     for(int i = ; i <= n; i++) {
         int item = lower_bound(hs+,hs++tot,i)-hs;
         int pos = min(i - , a[i]);
         ans += seg.query(, tot, item, tot, root[pos]);
     }

     printf("%lld\n",ans);
     return ;
 }
 /*
 */