Gym-10071A-Queries(树状数组)
链接:
https://vjudge.net/problem/Gym-100741A
题意:
Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my friend, a mathematician, thinks that it is very fun to play with a sequence of integer numbers. He writes the sequence in a row. If he wants he increases one number of the sequence, sometimes it is more interesting to decrease it (do you know why?..) And he likes to add the numbers in the interval [l;r]. But showing that he is really cool he adds only numbers which are equal some mod (modulo m).
Guess what he asked me, when he knew that I am a programmer? Yep, indeed, he asked me to write a program which could process these queries (n is the length of the sequence):
- p r It increases the number with index p by r. (, )
You have to output the number after the increase.
- p r It decreases the number with index p by r. (, ) You must not decrease the number if it would become negative.
You have to output the number after the decrease.
s l r mod You have to output the sum of numbers in the interval which are equal mod (modulo m). () ()
思路:
看半天没看懂题.
建10个树状数组即可, 对模m的每种情况分别统计.
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int INF = 1e9;
const int MAXN = 1e4+10;
LL A[MAXN], C[15][MAXN];
int n, m, q;
int Lowbit(int x)
{
return x&(-x);
}
void Add(int pos, int mod, LL val)
{
while (pos <= n)
{
C[mod][pos] += val;
pos += Lowbit(pos);
}
}
LL Query(int pos, int mod)
{
LL ans = 0;
while (pos > 0)
{
ans += C[mod][pos];
pos -= Lowbit(pos);
}
return ans;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1;i <= n;i++)
{
scanf("%lld", &A[i]);
Add(i, A[i]%m, A[i]);
}
scanf("%d", &q);
char opt[5];
int l, r, mod;
while (q--)
{
scanf("%s", opt);
if (opt[0] == 's')
{
scanf("%d%d%d", &l, &r, &mod);
printf("%lld\n", Query(r, mod)-Query(l-1, mod));
}
else if (opt[0] == '+')
{
scanf("%d%d", &l, &r);
Add(l, A[l]%m, -A[l]);
A[l] += r;
Add(l, A[l]%m, A[l]);
printf("%lld\n", A[l]);
}
else
{
scanf("%d%d", &l, &r);
Add(l, A[l]%m, -A[l]);
if (r <= A[l])
A[l] -= r;
Add(l, A[l]%m, A[l]);
printf("%lld\n", A[l]);
}
}
return 0;
}
Gym-10071A-Queries(树状数组)的更多相关文章
- GYM 100741A Queries(树状数组)
A. Queries time limit per test 0.25 seconds memory limit per test 64 megabytes input standard input ...
- Codeforces 369E Valera and Queries --树状数组+离线操作
题意:给一些线段,然后给m个查询,每次查询都给出一些点,问有多少条线段包含这个点集中的一个或多个点 解法:直接离线以点为基准和以线段为基准都不好处理,“正难则反”,我们试着求有多少线段是不包含某个查询 ...
- Gym - 101755G Underpalindromity (树状数组)
Let us call underpalindromity of array b of length k the minimal number of times one need to increme ...
- CF Gym 100463A (树状数组求逆序数)
题意:给你一个序列,和标准序列连线,求交点数. 题解:就是求逆序对个数,用树状数组优化就行了.具体过程就是按照顺序往树状数组了插点(根据点的大小),因为第i大的点应该排在第i位,插进去的时候他前面本该 ...
- Codeforces Round #216 (Div. 2) E. Valera and Queries 树状数组 离线处理
题意:n个线段[Li, Ri], m次询问, 每次询问由cnt个点组成,输出包含cnt个点中任意一个点的线段的总数. 由于是无修改的,所以我们首先应该往离线上想, 不过我是没想出来. 首先反着做,先求 ...
- GYM 101889F(树状数组)
bit扫描坐标套路题,注意有重复的点,莽WA了. const int maxn = 1e5 + 5; struct node { ll B, F, D; bool operator < (con ...
- gym 100589A queries on the Tree 树状数组 + 分块
题目传送门 题目大意: 给定一颗根节点为1的树,有两种操作,第一种操作是将与根节点距离为L的节点权值全部加上val,第二个操作是查询以x为根节点的子树的权重. 思路: 思考后发现,以dfs序建立树状数 ...
- Codeforces Gym 100114 H. Milestones 离线树状数组
H. Milestones Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Descripti ...
- Gym 101908C - Pizza Cutter - [树状数组]
题目链接:https://codeforces.com/gym/101908/problem/C 题意: 一块正方形披萨,有 $H$ 刀是横切的,$V$ 刀是竖切的,不存在大于等于三条直线交于一点.求 ...
- Codeforces Gym 100269F Flight Boarding Optimization 树状数组维护dp
Flight Boarding Optimization 题目连接: http://codeforces.com/gym/100269/attachments Description Peter is ...
随机推荐
- 5分钟了解图数据库Neo4j的使用
1.图数据库安装与配置 1.1安装与配置 配置path = %NEO4J_HOME%\bin 启动命令:neo4j console web访问:http://localhost:7474 1. ...
- postman关联及读取文件进行参数化
场景:登录后获取响应数据中的key.token..以便在接下来的接口调用.... 一.发送请求.查看响应 二.在Tests里使用响应的js代码来使其成为全局变量......... >>&g ...
- springboot 部署jar项目与自启动
1.项目路径 jar包路径:/usr/local/src/apps/govdata.jar 2.启动项目 nohup java -jar govdata.jar & 3.查看项目启动日志 ta ...
- Linu下安装与卸载MySQL数据库
卸载MySQL数据库,具体操作如下: (1)rpm -qa | grep -i mysql // 检查是否安装了MySQL的组件 (2)卸载前先关闭MySQL服务, a. b. (3)删除MySQL各 ...
- ABC136E Max GCD
Thinking about different ways of thinking. --- LzyRapx 题目 思路比较容易想到. Observations: 每次操作过后和不变. 枚举和的因子 ...
- C++学习 之 继承方式(笔记)
1.继承方式的分类 继承方式有公有继承,私有继承,保护继承.不同之处在于指定派生类的基类时使用的关键字不同:公有继承使用关键字public,私有继承使用关键字private,保护继承使用关键字prot ...
- P1828香甜的黄油
这是一道关于最短路的绿题. 题目给出一些农场,每个农场有奶牛,农场与农场之间存在边,要使所有奶牛到达其中一个农场的总距离最短,输出他们到达这个农场的距离.首先我想到了最小生成树,但我发现其实并不是,因 ...
- Design Support库中的控件
1.NavigationView滑动菜单 2.FloatIngActionButton悬浮按钮 3.Snackbar二次交互提示的按钮 4.Coordinatorlayout,监听子控件的各种事件(加 ...
- es6编译器(babel搭建)
1.安装Node.js,初始化项目 npm init -y 2.安装babel-cli(兼容至ie7) npm i @babel/core @babel/cli @babel/preset-env - ...
- 如何使用sftp下载Linux服务器上的文件到本地
下载Linux服务器上的文件到本地 Linux服务器上的操作 sftp xxxxx@jumper.xxxx.com 使用put命令进行文件上传,put app.log 本地操作 sftp xxxxx@ ...