KIDGIN7439的专栏

代码出自http://www.notonlysuccess.com/index.php/segment-tree-complete/ #include #include using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 5555; in

Tunnel Warfare Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15090Accepted Submission(s): 5957 Problem Description During the War of...

Atlantis Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20693Accepted Submission(s): 8258 Problem Description There are several ancie...

题目大意：给定n和k，输出1-k的所有置换按照字典序排列后的第n个置换，其中n由k个数Si给定 当i等于1的时候(k - 1)!就是后面k-1个数所有的置换的个数，然后前面乘si，所以第n个置换的第一个数字即si + 1，同理，可发现其实就是每次求第k大，然后删去该数字。 #include &lt;cstdio&gt; #include &lt;algorithm&gt; #includ...

G. Vasya and Maximum Profit time limit per test 3.5 seconds memory limit per test 256 megabytes input standard input output standard output Vasya got really tired of these credits (from probl...

B. Legacy time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Rick and his co-workers have made a new radioactive formula and a lot o...

题意:1 a:询问是不是有连续长度为a的空房间,有的话住进最左边 2 a b:将[a,a+b-1]的房间清空 思路:记录区间中最长的空房间 线段树操作:update:区间替换 query:询问满足条件的最左断点

最后的人的位置一定是固定的，所以倒序插入

因为是1到n，

注意到当一个数开方到1后就不再变了，每个数开方次数不会太多，所以进行更新时如果该区间的数字都是1了就直接return #include #include #include #define maxn 100100 __int64 sum[4*maxn]; void build(int l,int r,int o) { if(l==r) { scanf("%I6

以高为数轴，维护每个区间的剩余最大值，用线段树二分做 #include #define maxn 200100 int Max[4*maxn]; int h,w,n; int max(int a,int b) { if(a>b) return a; else return b; } void build(int l,int r,int o) { Max[o]=w;

线段树的区间修改 #include #define maxn 100100 int set[4*maxn],sum[4*maxn]; void push(int o,int m) { if(set[o]) { set[2*o]=set[2*o+1]=set[o]; sum[2*o]=(m-(m/2))*set[o]; sum[2*o+1]=(m/2)*

线段树的区间加减数，区间查询 #include #include using namespace std; #define LL __int64 const int maxn = 100100; LL sum[4*maxn]; LL add[4*maxn]; void build(int l,int r,int o) { add[o]=0; if(l==r) {

#include #include #define N 200200 int max[4*N]; int n,m; int A,B; int L,R; int MAX(int a,int b) { if(a>b) return a; else return b; } int MIN(int a,int b) { if(a<b) return a; else return b; } v

Kth number Time Limit: 15000/5000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12712Accepted Submission(s): 3871 Problem Description Give you a...