SuperMemo Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 11384 Accepted: 3572 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game…
SuperMemo Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 6841 Accepted: 2268 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game.…
题目连接 http://poj.org/problem?id=3580 SuperMemo Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A…
题目链接:http://poj.org/problem?id=3580 Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, ... An}. Then the h…
相应POJ题目:点击打开链接 SuperMemo Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 11309 Accepted: 3545 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a…
SuperMemo Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, ... An}. Then the host pe…
题意:让你维护一个序列,支持以下6种操作: ADD x y d: 第x个数到第y个数加d . REVERSE x y : 将区间[x,y]中的数翻转 . REVOLVE x y t :将区间[x,y]循环移位t次,如1 2 3 4 5 旋转2次后就变成4 5 1 2 3 . INSERT x p :在第x个数后面插入p . DELETE x :删除第x个数 . MIN x y : 查询区间[x,y]中的最小值 .思路:此题有反转区间和循环移位的操作,所以我们很容易可以想到用 splay FHQ_…
题意: 维护一个序列,支持如下几种操作: ADD x y D:将区间\([x,y]\)的数加上\(D\) REVERSE x y:翻转区间\([x,y]\) REVOLVE x y T:将区间\([x,y]\)向右循环平移\(T\)个长度 INSERT x P:在第\(x\)个元素后插入\(P\) DELETE x:删除第\(x\)个元素 QUERY x y:查询区间\([x,y]\)中的最小值 分析: ADD和REVERSE操作维护两个懒惰标记即可 REVOLVE操作本质还是CUT一段区间下来…
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 47944 Accepted: 14122 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…
A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3468 Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given nu…
A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3468 Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given nu…
Problem DescriptionAs we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A with n numbers. Then he makes m operations on it. There are three ty…
「模板」 线段树--区间乘 && 区间加 && 区间求和 原来的代码太恶心了,重贴一遍. #include <cstdio> int n,m; long long p; class SegmentTree { private: struct Node { int l,r; long long v,mul,add; Node *c[2]; Node(int l,int r):l(l),r(r),mul(1LL),add(0LL) { c[0]=c[1]=nullp…
题目链接:http://acm.uestc.edu.cn/#/problem/show/1344 区间加等差数列本质上就是区间修改区间查询,本来想用线段树做,结果这个题就是卡空间和时间……不得已学了区间修改区间查询的树状数组. #include<bits/stdc++.h> using namespace std; ; +; int a[maxn]; int N; int tree[maxn]; int tree2[maxn]; int lowbit(int x) { return x&…
题目描述 如题,已知一个数列,你需要进行下面三种操作: 1.将某区间每一个数乘上x 2.将某区间每一个数加上x 3.求出某区间每一个数的和 输入格式: 第一行包含三个整数N.M.P,分别表示该数列数字的个数.操作的总个数和模数. 第二行包含N个用空格分隔的整数,其中第i个数字表示数列第i项的初始值. 接下来M行每行包含3或4个整数,表示一个操作,具体如下: 操作1: 格式:1 x y k 含义:将区间[x,y]内每个数乘上k 操作2: 格式:2 x y k 含义:将区间[x,y]内每个数加上k…
