Codeforces_GYM_100741 A
http://codeforces.com/gym/100741/problem/A
A. Queries
0.25 seconds
64 megabytes
standard input
standard output
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). (
) (
)
The first line of each test case contains the number of elements of the sequence n and the number m. (1 ≤ n ≤ 10000) (1 ≤ m ≤ 10)
The second line contains n initial numbers of the sequence. (0 ≤ number ≤ 1000000000)
The third line of each test case contains the number of queries q (1 ≤ q ≤ 10000).
The following q lines contains the queries (one query per line).
Output q lines - the answers to the queries.
3 4
1 2 3
3
s 1 3 2
+ 2 1
- 1 2
2
3
1
m个 树状数组记录a[i]%m,得值。
#define LL long long
#include <iostream> using namespace std; const int MAXN();
const int N(+);
LL n,m,a[N],q,u,v,w; struct Tree
{
LL mm;
LL val[N];
#define lowbit(x) (x&(-x))
void Update(int now,int x)
{
for(;now<=mm;now+=lowbit(now)) val[now]+=x;
}
LL Query(int x)
{
LL ret=;
for(;x;x-=lowbit(x)) ret+=val[x];
return ret;
}
}tree[]; int main()
{
cin>>n>>m;
for(LL i=;i<m;i++) tree[i].mm=n;
for(LL i=;i<=n;i++)
{
cin>>a[i];
tree[a[i]%m].Update(i,a[i]);
}
cin>>q;
for(char ch[];q--;)
{
cin>>ch>>u>>v;
if(ch[]=='+')
{
tree[a[u]%m].Update(u,-a[u]);
a[u]+=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
} else
if(ch[]=='-')
{
if(a[u]<v) cout<<a[u]<<endl;
else
{
tree[a[u]%m].Update(u,-a[u]);
a[u]-=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
}
} else
if(ch[]=='s')
{
cin>>w;
cout<<tree[w].Query(v)-tree[w].Query(u-)<<endl;
}
}
return ;
}
Codeforces_GYM_100741 A的更多相关文章
随机推荐
- Centos7.6下安装Python3.7
前言 话说不会开发的运维不是一个好的DBA,所以我要开始学习python了,写博客记录一下我的学习过程,另外别欺负我新来的,那个每天更博的技术流ken是我哥. 不说了,时间宝贵,开整. 1.首先来看一 ...
- HDU——T 1251 统计难题
http://acm.hdu.edu.cn/showproblem.php?pid=1251 Time Limit: 4000/2000 MS (Java/Others) Memory Limi ...
- Windows7下修改pip源
以下列举三种方式的pip源配置: 1. 设置环境变量PIP_CONFIG_FILE指向pip.ini源配置文件,pip.ini文件内容如下: [global] index-url = http://m ...
- [Python] Working with file
Python allows you to open a file, do operations on it, and automatically close it afterwards using w ...
- (诡异Floyd&自环)MZ Training 2014 #15 E题(POJ 2240)
你们见过这么诡异的FLOYD吗? 先上题. [Description] 货币的汇率存在差异.比如,如果1美元购买0.5英镑,1英镑买10法郎.而1法国法郎买0.21美元.然后,通过转换货币,一个聪明的 ...
- 线段树 hdu3642 Get The Treasury
不得不说,这是一题很经典的体积并.. 然而还是debug了2个多小时... 首先思路:按z的大小排序. 然后相当于扫描面一样,,从体积的最下方向上方扫描,遇到这个面 就将相应的两条线增加到set中,或 ...
- vim 技巧之用宏命令批量处理文件
今天遇到了一种情况,就是我需要同时修改34个文件中的某些字符串的内容,如果一个个打开需改的话,那也太麻烦了.后来就想着能不能通过vim的宏命令来修改呢?现在就总结下关于宏在文件列表中的应用1.首先,我 ...
- POJ 1904 思路题
思路: 思路题 题目诡异地给了一组可行匹配 肯定有用啊-. 就把那组可行的解 女向男连一条有向边 如果男喜欢女 男向女连一条有向边 跑一边Tarjan就行了 (这个时候 环里的都能选 "增广 ...
- Linux 时区的修改
Linux 时区的修改 1. CentOS和Ubuntu的时区文件是/etc/localtime,但是在CentOS7以后localtime以及变成了一个链接文件 ``` [root@centos7 ...
- let---bash中用于计算的工具