A - Problem A. Ascending Rating

题意：给出n个数，给出区间长度m。对于每个区间，初始值的max为0，cnt为0.遇到一个a[i] > ans, 更新ans并且cnt++。计算

$A = \sum_{i = 1}^{i = n - m +1} (max \oplus i)$

$B = \sum_{i = 1}^{i = n - m +1} (cnt \oplus i)$

思路：单调队列，倒着扫一遍，对于每个区间的cnt就是队列的长度，扫一遍即可。

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = 1e7 + ;

 typedef long long ll;

 int t;

 int n,m,k;

 ll p, q, r, mod;

 ll arr[maxn];

 ll brr[maxn];

 int main()

 {

     scanf("%d", &t);

     while(t--)

     {

         scanf("%d %d %d %lld %lld %lld %lld", &n, &m, &k, &p, &q, &r, &mod);

         for(int i = ; i <= n; ++i)

         {

             if(i <= k) scanf("%lld", arr + i);

             else arr[i] = (p * arr[i - ] + q * i + r) % mod;

         }

         ll A = , B = ;

         int head = , tail = -;

         for(int i = n; i > (n - m + ); --i)

         {

             while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;

             brr[++tail] = i;

         }

         for(int i = n - m + ; i >= ; --i)

         {

             while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;

             brr[++tail] = i;

             while(brr[head] - i >= m) head++;

             A += arr[brr[head]] ^ i;

             B += (tail - head + ) ^ i;

         }

         printf("%lld %lld\n", A, B);

     }

     return ;

 }

B - Problem B. Cut The String

留坑。

C - Problem C. Dynamic Graph Matching

题意：给出n个点，两种操作，一种是加边，一种是减边，每次加一条或者减一条，每次操作后输出1, 2, ..., n/2 表示对应的数量的边，有几种取法，并且任意两个边不能相邻

思路：每加一条边，都可以转移状态 比如说 110100 到  111110 就是 dp[111110] += dp[110100]

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn =  << ;

 const int MOD = 1e9 + ;

 int t, n, m;

 ll dp[maxn];

 ll ans[];

 inline int cal(int x)

 {

     int cnt = ;

     while(x)

     {

         if(x & ) cnt++;

         x >>= ;

     }

     return cnt;

 }

 int main()

 {

     scanf("%d", &t);

     while(t--)

     {

         memset(ans, , sizeof ans);

         memset(dp, , sizeof dp);

         dp[] = ;

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

         while(m--)

         {

             char op[];

             int u, v;

             scanf("%s %d %d", op, &u, &v);

             --u,--v;

             for(int s =  ; s < ( << n); ++s)

             {

                 if((s & ( << u)) || (s & ( << v))) continue;

                 int S = s | ( << u);

                 S |= ( << v);

                 if(op[] == '+')

                 {

                     dp[S] = (dp[S] + dp[s]) % MOD;

                     ans[cal(S)] = (ans[cal(S)] + dp[s]) % MOD;

                 }

                 else if(op[] == '-')

                 {

                     dp[S] = (dp[S] - dp[s] + MOD) % MOD;

                     ans[cal(S)] = (ans[cal(S)] - dp[s] + MOD) % MOD;

                 }

             }

             for(int i = ; i <= n; i += )

             {

                 printf("%lld%c", ans[i], " \n"[i == n]);

             }

         }

     }

     return ;

 }

D - Problem D. Euler Function

打表找规律

 #include <bits/stdc++.h>

 using namespace std;

 #define N 100010

 inline int gcd(int a, int b)

 {

     return b ==  ? a : gcd(b, a % b);

 }

 inline void Init()

 {

     for (int i = ; i <= ; ++i)

     {

         int res = ;

         for (int j = ; j < i; ++j)

             res += (gcd(i, j) == );

         printf("%d %d\n", i, res);

     }

 }

 int t, n;

 int main()

 {

     scanf("%d", &t);

     while (t--)

     {

         scanf("%d", &n);

         if (n == ) puts("");

         else printf("%d\n", n + );

     }

     return ;

 }

E - Problem E. Find The Submatrix

留坑，

F - Problem F. Grab The Tree

题意：一棵树，每个点有点权，小Q可以选择若干个点，并且任意两个点之间没有边，小T就是剩下的所有点，然后两个人的值就是拥有的点异或起来，求小Q赢还是输还是平局

思路：显然，只有赢或者平局的情况，只要考虑是否有平局情况就可以，当所有点异或起来为0便是平局

 #include <bits/stdc++.h>

 using namespace std;

 #define N 100010

 int t, n;

 int arr[N];

 int main()

 {

     scanf("%d", &t);

     while (t--)

     {

         scanf("%d", &n);

         int res = ;

         for (int i = ; i <= n; ++i)

         {

             scanf("%d", arr + i);

             res ^= arr[i];

         }

         for (int i = , u, v; i < n; ++i) scanf("%d%d", &u, &v);

         if (res == )

         {

             puts("D");

             continue;

         }

         puts("Q");

     }

     return ;

 }

G - Problem G. Interstellar Travel

题意：给出n个点，要从$i -> j$ 当且仅当 $x_i < x_j$ 花费为 $x_i \cdot y_j - x_j \cdot y_i$ 求从$1 -> n$ 求权值最小如果多解输出字典序最小的解

思路：可以发现权值就是叉积，求个凸包，然后考虑在一条线上的情况

 #include <bits/stdc++.h>

 using namespace std;

 #define N 200010

 const double eps = 1e-;

 inline int sgn(double x)

 {

     if (fabs(x) < eps) return ;

     if (x < ) return -;

     else return ;

 }

 struct Point

 {

     double x, y; int id;

     inline Point () {}

     inline Point(double x, double y) : x(x), y(y) {}

     inline void scan(int _id)

     {

         id = _id;

         scanf("%lf%lf", &x, &y);

     }

     inline bool operator == (const Point &r) const { return sgn(x - r.x) ==  && sgn(y - r.y) == ; }

     inline bool operator < (const Point &r) const { return x < r.x || x == r.x && y < r.y || x == r.x && y == r.y && id < r.id; }

     inline Point operator + (const Point &r) const { return Point(x + r.x, y + r.y); }

     inline Point operator - (const Point &r) const { return Point(x - r.x, y - r.y); }

     inline double operator ^ (const Point &r) const { return x * r.y - y * r.x; }

     inline double distance(const Point &r) const { return hypot(x - r.x, y - r.y); }

 };

 struct Polygon

 {

     int n;

     Point p[N];

     struct cmp

     {

         Point p;

         inline cmp(const Point &p0) { p = p0; }

         inline bool operator () (const Point &aa, const Point &bb)

         {

             Point a = aa, b = bb;

             int d = sgn((a - p) ^ (b - p));

             if (d == )

             {

                 return sgn(a.distance(p) - b.distance(p)) < ;

             }

             return d < ;

         }

     };

     inline void norm()

     {

         Point mi = p[];

         for (int i = ; i < n; ++i) mi = min(mi, p[i]);

         sort(p, p + n, cmp(mi));

     }

     inline void Graham(Polygon &convex)

     {

         sort(p + , p + n - );

         int cnt = ;

         for (int i = ; i < n; ++i)

         {

             if (!(p[i] == p[i - ]))

                 p[cnt++] = p[i];

         }

         n = cnt;

         norm();

         int &top = convex.n;

         top = ;

         if (n == )

         {

             top = ;

             convex.p[] = p[];

             convex.p[] = p[];

             return;

         }

         if(n == )

         {

             top = ;

             convex.p[] = p[];

             convex.p[] = p[];

             convex.p[] = p[];

             return ;

         }

         convex.p[] = p[];

         convex.p[] = p[];

         top = ;

         for (int i = ; i < n; ++i)

         {

             while (top >  && sgn((convex.p[top - ] - convex.p[top - ]) ^ (p[i] - convex.p[top - ])) >= )

             {

                 if(sgn((convex.p[top - ] - convex.p[top - ]) ^ (p[i] - convex.p[top - ])) == )

                 {

                     if(p[i].id < convex.p[top - ].id) --top;

                     else break;

                 }

                 else

                 {

                     --top;

                 }

             }

             convex.p[top++] = p[i];

         }

         if (convex.n ==  && (convex.p[] == convex.p[])) --convex.n;

     }

 }arr, ans;

 int t, n;

 int main()

 {

     scanf("%d", &t);

     while (t--)

     {

         scanf("%d", &n); arr.n = n;

     //    cout << n << endl;

         for (int i = ; i < n; ++i) arr.p[i].scan(i + );

         arr.Graham(ans);

         for (int i = ; i < ans.n; ++i) printf("%d%c", ans.p[i].id, " \n"[i == ans.n - ]);

     }

     return ;

 }

H - Problem H. Monster Hunter

留坑。

I - Problem I. Random Sequence

留坑。

J - Problem J. Rectangle Radar Scanner

留坑。

K - Problem K. Transport Construction

留坑。

L - Problem L. Visual Cube

按题意模拟即可，分块解决

 #include <bits/stdc++.h>

 using namespace std;

 int t, a, b, c, n, m;

 char ans[][];

 int main()

 {

     scanf("%d", &t);

     while (t--)

     {

         memset(ans, , sizeof ans);

         scanf("%d%d%d", &a, &b, &c);

         n = (b + c) *  + ;

         m = (a + b) *  + ;

         for (int i = n, cnt = ; cnt <= c; ++cnt, i -= )

         {

             for (int j = ; j <=  * a; j += )

                 ans[i][j] = '+', ans[i][j + ] = '-';

         }

         for (int i = n - , cnt = ; cnt <= c; ++cnt, i -= )

         {

             for (int j = ; j <=  * a; j += )

                 ans[i][j] = '|';

         }

         for (int i =  * b + , tmp = , cnt = ; cnt <= b + ; ++cnt, i -=, tmp += )

         {

             for (int j =  + tmp, cntt = ; cntt <= a; ++cntt, j += )

                ans[i][j] = '+', ans[i][j + ] = '-';

         }

         for (int i =  * b, tmp = , cnt = ; cnt <= b; ++cnt, tmp += , i -= )

         {

             for (int j =  + tmp, cntt = ; cntt <= a; ++cntt, j += )

                 ans[i][j] = '/';

         }

         for (int j = m, cntt = , tmp = ; cntt <= b + ; ++cntt, j -= , tmp += )

         {

             for (int i =  + tmp, cnt = ; cnt <= c + ; ++cnt, i += )

             {

                 ans[i][j] = '+';

                 if (cnt <= c) ans[i + ][j] = '|';

             }

         }

         for (int j = m - , tmp = , cntt = ; cntt <= b; ++cntt, tmp += , j -= )

         {

             for (int i =  + tmp, cnt = ; cnt <= c + ; ++cnt, i += )

                 ans[i][j] = '/';

         }

         for (int i = ; i <= n; ++i)

         {

             for (int j = ; j <= m; ++j)

                 if (!ans[i][j]) ans[i][j] = '.';

             ans[i][m + ] = ;

             printf("%s\n", ans[i] + );

         }

     }

     return ;

 }

M - Problem M. Walking Plan

题意：给出一张图，询问从$u -> v 至少经过k条路的最少花费$

思路：因为$k <= 10000$ 那么我们可以拆成 $k = x * 100 + y$ 然后考虑分块

$G[k][i][j] 表示 i -> j 至少经过k条路$

$dp[k][i][j] 表示 i -> j 至少经过 k * 100 条路$

然后查询的时候枚举中间点

 #include <bits/stdc++.h>

 using namespace std; 

 #define N 210

 #define INFLL 0x3f3f3f3f3f3f3f3f

 #define ll long long

 int t, n, m, q;

 ll G[N][][], dp[N][][];

 inline void Init()

 {

     memset(G, 0x3f, sizeof G);

     memset(dp, 0x3f, sizeof dp);

 }

 inline void Floyd()

 {

     for (int l = ; l <= ; ++l)

         for (int k = ; k <= n; ++k)

             for (int i = ; i <= n; ++i)

                 for (int j = ; j <= n; ++j)

                     G[l][i][j] = min(G[l][i][j], G[l - ][i][k] + G[][k][j]);

     for (int i = ; i <= n; ++i)

         for (int j = ; j <= n; ++j)

             for (int l = ; l >= ; --l)

                 G[l][i][j] = min(G[l][i][j], G[l + ][i][j]); // at least K roads;

     for (int i = ; i <= n; ++i)

         for (int j = ; j <= n; ++j)

             dp[][i][j] = G[][i][j];

     for (int l = ; l <= ; ++l)

         for (int k = ; k <= n; ++k)

             for (int i = ; i <= n; ++i)

                 for (int j = ; j <= n; ++j)

                     dp[l][i][j] = min(dp[l][i][j], dp[l - ][i][k] + dp[][k][j]);

 }

 inline void Run()

 {

     scanf("%d", &t);

     while (Init(), t--)

     {

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

         for (int i = , u, v, w; i <= m; ++i)

         {

             scanf("%d%d%d", &u, &v, &w);

             G[][u][v] = min(G[][u][v], (ll)w);

         }

         Floyd(); scanf("%d", &q);

         for (int i = , u, v, k; i <= q; ++i)

         {

             scanf("%d%d%d", &u, &v, &k);

             int unit =  floor (k * 1.0 / ), remind = k - unit * ;

             ll ans = INFLL;

             if (k <= ) ans = G[k][u][v];

             else

             {

                 for (int j = ; j <= n; ++j)

                     ans = min(ans, dp[unit][u][j] + G[remind][j][v]);

             }

             if (k >  && remind == ) ans = min(ans, dp[unit][u][v]);

             printf("%lld\n", ans >= INFLL ? - : ans);

         }

     }

 }

 int main()

 {

     #ifdef LOCAL

         freopen("Test.in", "r", stdin);

     #endif  

     Run();

     return ;

 }

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