Mobile phones POJ - 1195 二维树状数组求和
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.
Input
The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30
Output
Sample Input
0 4
1 1 2 3
2 0 0 2 2
1 1 1 2
1 1 2 -1
2 1 1 2 3
3
Sample Output
3
4 板子题
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define fuck(x) cout<<"["<<x<<"]"<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define lowbit(x) x&-x
#pragma comment (linker,"/STACK:102400000,102400000") using namespace std;
typedef long long LL ;
const int maxn = 2e3 + ;
int n, k, c[maxn][maxn];
void updata(int x, int y, int z) {
for (int i = x ; i <= n ; i += lowbit(i))
for (int j = y ; j <= n ; j += lowbit(j))
c[i][j] += z;
}
int sum(int x, int y) {
int ret = ;
for (int i = x ; i > ; i -= lowbit(i))
for (int j = y ; j > ; j -= lowbit(j))
ret += c[i][j];
return ret;
}
int main() {
scanf("%d%d", &k, &n);
while() {
scanf("%d", &k);
if (k == ) {
int x, y, z;
sfff(x, y, z);
x++, y++;
updata(x, y, z);
}
if (k == ) {
int x1,y1,x2,y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
x1++, y1++, x2++, y2++;
int ans = sum(x2, y2) - sum(x1 - , y2) - sum(x2, y1 - ) + sum(x1 - , y1 - );
printf("%d\n", ans);
}
if (k == ) break;
}
return ;
}
Mobile phones POJ - 1195 二维树状数组求和的更多相关文章
- POJ 1195 二维树状数组
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 18489 Accepted: 8558 De ...
- poj 2029 二维树状数组
思路:简单树状数组 #include<map> #include<set> #include<cmath> #include<queue> #inclu ...
- poj 3378 二维树状数组
思路:直接用long long 保存会WA.用下高精度加法就行了. #include<map> #include<set> #include<cmath> #inc ...
- poj 2155 (二维树状数组 区间修改 求某点值)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 33682 Accepted: 12194 Descript ...
- HihoCoder1336 Matrix Sum(二维树状数组求和)
时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 You are given an N × N matrix. At the beginning every element ...
- POJ 1195 Mobile phones (二维树状数组)
Description Suppose that the fourth generation mobile phone base stations in the Tampere area operat ...
- poj 1195:Mobile phones(二维树状数组,矩阵求和)
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 14489 Accepted: 6735 De ...
- (简单) POJ 1195 Mobile phones,二维树状数组。
Description Suppose that the fourth generation mobile phone base stations in the Tampere area operat ...
- POJ 1195:Mobile phones 二维树状数组
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 16893 Accepted: 7789 De ...
随机推荐
- 剑指offer-栈的压入弹出序列21
题目描述 输入两个整数序列,第一个序列表示栈的压入顺序,请判断第二个序列是否可能为该栈的弹出顺序.假设压入栈的所有数字均不相等.例如序列1,2,3,4,5是某栈的压入顺序,序列4,5,3,2,1是该压 ...
- day-17 L1和L2正则化的tensorflow示例
机器学习中几乎都可以看到损失函数后面会添加一个额外项,常用的额外项一般有两种,一般英文称作ℓ1-norm和ℓ2-norm,中文称作L1正则化和L2正则化,或者L1范数和L2范数.L2范数也被称为权重衰 ...
- 20145214实验三 敏捷开发与XP实践
20145214实验三 敏捷开发与XP实践 XP准则 沟通 :XP认为项目成员之间的沟通是项目成功的关键,并把沟通看作项目中间协调与合作的主要推动因素. 简单 :XP假定未来不能可靠地预测,在现在考虑 ...
- c#积累之测试
初来上班,免不了看别人代码.快速搞懂别人代码是我现在受到的一大挑战.寻摸着规律,发现一边进行调试,一边进行行行注释的逻辑判断不失为一种妙招. c#调试用的是vs2012.f11键和f10和f5键的应用 ...
- JavaScript初探系列之Ajax应用
一 什么是Ajax Ajax是(Asynchronous JavaScript And XML)是异步的JavaScript和xml.也就是异步请求更新技术.Ajax是一种对现有技术的一种新的应用,不 ...
- android入门 — ListView
ListView主要是用来解决大量数据展示的问题,它的用途很广泛,几乎所有的app都会用到,比如说知乎.今日头条.微博.通讯录等. ListView允许用户通过上下滑动的方式将屏幕外的数据滚动到屏幕中 ...
- Swift-可选值(Optional)讲解
前提:Swift中有规定:对象中的任何属性在创建时,都必须要有明确的初始化值 1.定义可选类型 方式一:常规方式(不常用) var name : Optional<String> = ni ...
- lol人物模型提取(八)
今天顺风终于把包裹送到了北航新主楼自提柜,怀着激动喜悦的心情,我小心翼翼地将其取回. 到了晚上,是时候解开佐伊的封印了! 开了个小口,发现里面包得还挺严实的. 去掉了纸盒,里面还有一层 ...
- IPython 4.0发布:Jupyter和IPython分离后的首个版本
IPython 4.0日前正式发布,这是IPython分离成IPython和Jupyter后的第一个重要版本. 更新Jupyter的快捷方式是: pip install --upgrade jupyt ...
- 【转】log4j.properties文件的配置
一.前言 log4j使用的还是比较多的,但是对于其配置又很难描述清楚要怎么配置,说明我自己对于log4j的配置并不是非常熟悉,所以在网上找了一篇详尽的 博文转载,在此非常感谢原文作者的辛苦付出,如有需 ...