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#include #include using namespace std;struct student { int need, pos;};int main(){ int n, m; student s; while (cin >> n >> m) { queue q; for (int i = 1; i <= n; ++i) { cin >> s.nee

Description现在请求你维护一个数列，要求提供以下两种操作： 1、 查询操作。语法：Q L 功能：查询当前数列中末尾L个数中的最大的数，并输出这个数的值。限制：L不超过当前数列的长度。 2、 插入操作。语法：A n 功能：将n加上t，其中t是最近一次查询操作的答案（如果还未执行过查询操作，则t=0)，并将所得结果对一个固定的常数D取模，将所得答案插入到数列的末尾。限制：n是非负整数

/*线段树*/#include #include #include #include using namespace std;const int maxn = 100000 + 5;struct node { int s, e, id; node() {}; bool operator <(const node &a)const { if (a.s == s) retu

#include #include #include using namespace std;const int maxn = 50000 + 5;struct node { int l, r, sum;};node tree[maxn * 4];int t, cas = 0, n, ans;char s[10];void build(int l, int r, in

/*线段树 区间维护当前可插入的个数*/#include #include #include using namespace std;const int maxn = 200000 + 5;struct node { int l, r, sum;};struct person { int pos, val;};node tree[maxn << 2];person

#include using namespace std;const int maxn = 8000 + 5;struct node { int l, r, len;};node tree[maxn * 4];int n;int ans[maxn], pre[maxn];void build(int l, int r, int root){ tree[root].l

/*线段树 单点更新 +　寻找区间最值*/#include #include #include #include using namespace std;const int maxn = 200000 + 5;struct node { int l, r, Max, score;};node tree[maxn * 4];int n, m, ans;void bu

/*Interval Tree*/#include #include #include using namespace std;struct node { int l, r, val;};int n, x, y;node tree[32005 * 4];int level[32005];void creat(int l, int r, int root){ tre

#include #include #include #include using namespace std;struct node { int l, r; int MIN, MAX, num;};const int maxn = 500010;const int INF = 100000000;node tree[maxn << 2];int n, q;int

/*并查集*/#include #include using namespace std;const int MAXN = 2005;int father[MAXN], Rank[MAXN];int cas = 0, t;int n, m;bool flag;void init(){ for (int i = 0; i < n; ++i) { father[i] =

/*并查集*/#include #include using namespace std;const int MAXN = 30005;int father[MAXN], Rank[MAXN], up[MAXN];int p;void init(){ for (int i = 0; i < MAXN; ++i) { father[i] = i; Rank[i] =

/*并查集*/#include using namespace std;const int MAXN = 50005;int father[MAXN];int n, m, cas = 0, sum;void init(){ for (int i = 0; i < n; ++i) { father[i] = i; } return;}int getfather(i

/*并查集*/#include #include #include using namespace std;const int MAXN = 1005;int father[MAXN];int n, d;struct node { double x, y; int flag;};node parent[MAXN];void init(){ for (int i

#include using namespace std;const int MAXN = 30005;int n, m;int father[MAXN], RANK[MAXN];void init() { for (int i = 0; i < n; ++i) { father[i] = i; RANK[i] = 1; } return;}int find_r

Description 在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为可见的,否则Li为被覆盖的.    例如,对于直线:    L1:y=x; L2:y=-x; L3:y=0    则L1和L2是可见的,L3是被覆盖的.    给出n条直线,表示成y=Ax+B的形式(|A|,|B|Input