Remainder Problem(分块) Educational Codeforces Round 71 (Rated for Div. 2)
引用:https://blog.csdn.net/qq_41879343/article/details/100565031

下面代码写错了,注意要上面这种。查:2 800 0,下面代码就错了。
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
#define fo(a,b,c) for(register int a=b;a<=c;++a)
#define fr(a,b,c) for(register int a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef long long ll;
void swapp(int &a,int &b);
double fabss(double a);
int maxx(int a,int b);
int minn(int a,int b);
int Del_bit_1(int n);
int lowbit(int n);
int abss(int a);
//const long long INF=(1LL<<60);
const double E=2.718281828;
const double PI=acos(-1.0);
const int inf=(<<);
const double ESP=1e-;
const int mod=(int)1e9+;
const int N=(int)1e6+; ll a[N];
ll ans[][]; int main()
{
int n;
sc("%d",&n);
fo(n_,,n)
{
int judge,x,y;
sc("%d%d%d",&judge,&x,&y);
if(judge==)
{
a[x]+=y;
for(int i=;i<=;++i)
ans[x%i][i]+=y;
}
else
{
ll Ans=;
if(x<=)
Ans+=ans[y][x];
else
for(int i=;i<=;i+=x)
Ans+=a[i+y-];
pr("%lld\n",Ans);
}
}
return ;
} /**************************************************************************************/ int maxx(int a,int b)
{
return a>b?a:b;
} void swapp(int &a,int &b)
{
a^=b^=a^=b;
} int lowbit(int n)
{
return n&(-n);
} int Del_bit_1(int n)
{
return n&(n-);
} int abss(int a)
{
return a>?a:-a;
} double fabss(double a)
{
return a>?a:-a;
} int minn(int a,int b)
{
return a<b?a:b;
}
Remainder Problem(分块) Educational Codeforces Round 71 (Rated for Div. 2)的更多相关文章
- Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块
Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块 [Problem Description] ...
- Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题
Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题 [Problem Description] 总共两次询 ...
- Educational Codeforces Round 71 (Rated for Div. 2)
传送门 A.There Are Two Types Of Burgers 签到. B.Square Filling 签到 C.Gas Pipeline 每个位置只有"高.低"两种状 ...
- Educational Codeforces Round 71 (Rated for Div. 2) Solution
A. There Are Two Types Of Burgers 题意: 给一些面包,鸡肉,牛肉,你可以做成鸡肉汉堡或者牛肉汉堡并卖掉 一个鸡肉汉堡需要两个面包和一个鸡肉,牛肉汉堡需要两个面包和一个 ...
- Educational Codeforces Round 71 (Rated for Div. 2)E. XOR Guessing
一道容斥题 如果直接做就是找到所有出现过递减的不同排列,当时硬钢到自闭,然后在凯妹毁人不倦的教导下想到可以容斥做,就是:所有的排列设为a,只考虑第一个非递减设为b,第二个非递减设为c+两个都非递减的情 ...
- Educational Codeforces Round 71 (Rated for Div. 2) E XOR Guessing (二进制分组,交互)
E. XOR Guessing time limit per test1 second memory limit per test256 megabytes inputstandard input o ...
- [暴力] Educational Codeforces Round 71 (Rated for Div. 2) B. Square Filling (1207B)
题目:http://codeforces.com/contest/1207/problem/B B. Square Filling time limit per test 1 second mem ...
- [贪心,dp] Educational Codeforces Round 71 (Rated for Div. 2) C. Gas Pipeline (1207C)
题目:http://codeforces.com/contest/1207/problem/C C. Gas Pipeline time limit per test 2 seconds memo ...
- XOR Guessing(交互题+思维)Educational Codeforces Round 71 (Rated for Div. 2)
题意:https://codeforc.es/contest/1207/problem/E 答案guessing(0~2^14-1) 有两次机会,内次必须输出不同的100个数,每次系统会随机挑一个你给 ...
随机推荐
- google中select添加onclick
有下拉跳转框如下所示: <select name="page" size="1" > <option onclick="refurb ...
- 2016 NEERC, Northern Subregional Contest G.Gangsters in Central City(LCA)
G.Gangsters in Central City 题意:一棵树,节点1为根,是水源.水顺着边流至叶子.该树的每个叶子上有房子.有q个询问,一种为房子u被强盗入侵,另一种为强盗撤离房子u.对于每个 ...
- 预处理、const、static与sizeof-用宏定义得到一个字的高位和低位字节
1:代码如下: #define WORD_LO(XXX) ((byte) (word)(XXX) & 255) #define WORD_HI(XXX) ((byte) (word)(XXX) ...
- gradle添加阿里云maven库
用gradle构建spring项目,才发现gradle要添加阿里云maven库和maven不太一样 链接:https://www.cnblogs.com/SiriYang/p/10638365.htm ...
- flutter常用内置动画组件
文章目录 AnimatedContainer AnimatedCrossFade Hero AnimatedBuilder DecoratedBoxTransition FadeTransition ...
- How to use reminder feature of the outlook
https://support.office.com/en-us/article/set-or-remove-reminders-7a992377-ca93-4ddd-a711-851ef359792 ...
- Flume-Spooling Directory Source 监控目录下多个新文件
使用 Flume 监听整个目录的文件,并上传至 HDFS. 一.创建配置文件 flume-dir-hdfs.conf https://flume.apache.org/FlumeUserGuide.h ...
- 对 Python 迭代的深入研究
在程序设计中,通常会有 loop.iterate.traversal 和 recursion 等概念,他们各自的含义如下: 循环(loop),指的是在满足条件的情况下,重复执行同一段代码.比如 Pyt ...
- Golang 空指针nil的方法和数据成员
golang中,有一个特殊的指针值nil. 如何使用nil没有方法和成员变量呢? 下面来看下具体例子. 程序中,定义结构体类型Plane, 将Plane类型的指针作为函数的参数,然后传入nil作为实参 ...
- easyUI之Dialog(对话框窗口)
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <hea ...