Prime Query Time Limit: 1 Second      Memory Limit: 196608 KB You are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on the given sequence. Here are the operations: A v l, add the value v to element with i…
Prime Query Time Limit: 1 Second      Memory Limit: 196608 KB You are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on the given sequence. Here are the operations: A v l, add the value v to element with i…
Bob wants to pour water Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge There is a huge cubiod house with infinite height. And there are some spheres and some cuboids in the house. They do not intersect with others and the house.…
Market Time Limit: 2 Seconds      Memory Limit: 65536 KB There's a fruit market in Byteland. The salesmen there only sell apples. There are n salesmen in the fruit market and the i-th salesman will sell at most wi apples. Every salesman has an immedi…
Number Game Time Limit: 2 Seconds      Memory Limit: 65536 KB The bored Bob is playing a number game. In the beginning, there are n numbers. For each turn, Bob will take out two numbers from the remaining numbers, and get the product of them. There i…
Cake Time Limit: 4 Seconds      Memory Limit: 65536 KB Alice and Bob like eating cake very much. One day, Alice and Bob went to a bakery and bought many cakes. Now we know that they have bought n cakes in the bakery. Both of them like delicious cakes…
Ant Time Limit: 1 Second      Memory Limit: 32768 KB There is an ant named Alice. Alice likes going hiking very much. Today, she wants to climb a cuboid. The length of cuboid's longest edge is n, and the other edges are all positive integers. Alice's…
题目传送门 /* 题意:n个时刻点,m次时光穿梭,告诉的起点和终点,q次询问,每次询问t时刻t之前有多少时刻点是可以通过两种不同的路径到达 思维:对于当前p时间,从现在到未来穿越到过去的是有效的值,排个序,从大到小询问,那么之前添加的穿越点都是有效的, 用multiset保存.比赛时想到了排序,但是无法用线段树实现查询,stl大法好! */ #include <cstdio> #include <algorithm> #include <cstring> #includ…
题目链接:ZOJ 2706 Thermal Death of the Universe (线段树) 题意:n个数.m个操作. 每一个操作(a,b)表示(a,b)全部值更新为这个区间的平均数:1.当前的数列总和小于等于原数列总和.取平均值的上界,反之.取下界. 注意有负数的情况. AC代码: #include<stdio.h> #include <math.h> #define LL long long #define lson l,m,rt<<1 #define rso…
Solution: 根据树的遍历道的时间给树的节点编号,记录下进入节点和退出节点的时间.这个时间区间覆盖了这个节点的所有子树,可以当做连续的区间利用线段树进行操作. /* 线段树 */ #pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…