A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 57666   Accepted: 17546
Case Time Limit: 2000MS

Description

You have N integers, A1A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is
to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.

The second line contains N numbers, the initial values of A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.

"Q a b" means querying the sum of AaAa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

2014-9-4 16:25:07更新

#include <stdio.h>
#include <string.h>
#define maxn 100002
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll; struct Node{
ll lazy, sum;
} tree[maxn << 2]; void pushUp(int rt){
tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
} void pushDown(int l, int r, int rt)
{
int mid = (l + r) >> 1;
tree[rt << 1].sum += (mid - l + 1) * tree[rt].lazy;
tree[rt << 1 | 1].sum += (r - mid) * tree[rt].lazy;
tree[rt << 1].lazy += tree[rt].lazy;
tree[rt << 1 | 1].lazy += tree[rt].lazy;
tree[rt].lazy = 0;
} void build(int l, int r, int rt)
{
tree[rt].lazy = 0;
if(l == r){
scanf("%lld", &tree[rt].sum);
return;
}
int mid = (l + r) >> 1;
build(lson); build(rson);
pushUp(rt);
} void update(int left, int right, ll val, int l, int r, int rt)
{
if(left == l && right == r){
tree[rt].sum += (r - l + 1) * val;
tree[rt].lazy += val; return;
}
int mid = (l + r) >> 1;
if(tree[rt].lazy) pushDown(l, r, rt);
if(right <= mid) update(left, right, val, lson);
else if(left > mid) update(left, right, val, rson);
else {
update(left, mid, val, lson);
update(mid + 1, right, val, rson);
}
pushUp(rt);
} ll query(int left, int right, int l, int r, int rt)
{
if(left == l && right == r) return tree[rt].sum;
int mid = (l + r) >> 1;
if(tree[rt].lazy) pushDown(l, r, rt);
if(right <= mid) return query(left, right, lson);
else if(left > mid) return query(left, right, rson);
return query(left, mid, lson) + query(mid + 1, right, rson);
} int main()
{
int n, m, i, a, b;
char com[2];
ll c;
while(scanf("%d%d", &n, &m) == 2){
build(1, n, 1);
while(m--){
scanf("%s%d%d", com, &a, &b);
if(com[0] == 'Q')
printf("%lld\n", query(a, b, 1, n, 1));
else{
scanf("%lld", &c);
update(a, b, c, 1, n, 1);
}
}
}
return 0;
}

RE了三次,发现是query函数内的pushdown函数位置放错了。

改了下,最终A掉了。

//#define DEBUG
#include <stdio.h>
#define maxn 100002
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1 long long tree[maxn << 2], arr[maxn], lazy[maxn << 2]; void pushUp(int rt)
{
tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
} void pushDown(int l, int r, int rt)
{
int mid = (l + r) >> 1; tree[rt << 1] += (mid - l + 1) * lazy[rt];
tree[rt << 1 | 1] += (r - mid) * lazy[rt]; lazy[rt << 1] += lazy[rt];
lazy[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
} void build(int l, int r, int rt)
{
lazy[rt] = 0;
if(l == r){
tree[rt] = arr[r]; return;
} int mid = (l + r) >> 1;
build(lson);
build(rson); pushUp(rt);
} void update(int left, int right, long long val, int l, int r, int rt)
{
if(left == l && right == r){
lazy[rt] += val; tree[rt] += val * (r - l + 1); return;
} //include l == r if(lazy[rt]) pushDown(l, r, rt); int mid = (l + r) >> 1;
if(right <= mid) update(left, right, val, lson);
else if(left > mid) update(left, right, val, rson);
else{
update(left, mid, val, lson);
update(mid + 1, right, val, rson);
} pushUp(rt);
} long long query(int left, int right, int l, int r, int rt)
{
if(left == l && right == r)
return tree[rt]; if(lazy[rt]) pushDown(l, r, rt); int mid = (l + r) >> 1;
if(right <= mid){
return query(left, right, lson);
}else if(left > mid){
return query(left, right, rson);
}else{
return query(left, mid, lson) + query(mid + 1, right, rson);
}
} int main()
{
#ifdef DEBUG
freopen("../stdin.txt", "r", stdin);
freopen("../stdout.txt", "w", stdout);
#endif int n, q, i, a, b;
long long c;
char com[2];
while(scanf("%d%d", &n, &q) == 2){
for(i = 1; i <= n; ++i)
scanf("%lld", arr + i);
build(1, n, 1); while(q--){
scanf("%s%d%d", com, &a, &b);
if(com[0] == 'C'){
scanf("%lld", &c);
update(a, b, c, 1, n, 1);
}else printf("%lld\n", query(a, b, 1, n, 1));
}
}
return 0;
}

版权声明:本文博客原创文章,博客,未经同意,不得转载。

POJ3468 A Simple Problem with Integers 【段树】+【成段更新】的更多相关文章

  1. POJ 3468 A Simple Problem with Integers(线段树 成段增减+区间求和)

    A Simple Problem with Integers [题目链接]A Simple Problem with Integers [题目类型]线段树 成段增减+区间求和 &题解: 线段树 ...

  2. 【POJ】3468 A Simple Problem with Integers ——线段树 成段更新 懒惰标记

    A Simple Problem with Integers Time Limit:5000MS   Memory Limit:131072K Case Time Limit:2000MS Descr ...

  3. POJ3468_A Simple Problem with Integers(线段树/成段更新)

    解题报告 题意: 略 思路: 线段树成段更新,区间求和. #include <iostream> #include <cstring> #include <cstdio& ...

  4. POJ 3468 A Simple Problem with Integers (线段树成段更新)

    题目链接:http://poj.org/problem?id=3468 题意就是给你一组数据,成段累加,成段查询. 很久之前做的,复习了一下成段更新,就是在单点更新基础上多了一个懒惰标记变量.upda ...

  5. POJ-3468 A Simple Problem with Integers (区间求和,成段加减)

    You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of op ...

  6. poj3468 A Simple Problem with Integers (线段树区间最大值)

    A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92127   ...

  7. poj3468 A Simple Problem with Integers (树状数组做法)

    题目传送门 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 1 ...

  8. A Simple Problem with Integers(线段树,区间更新)

    A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 83822   ...

  9. poj3468 A Simple Problem with Integers(线段树模板 功能:区间增减,区间求和)

    转载请注明出处:http://blog.csdn.net/u012860063 Description You have N integers, A1, A2, ... , AN. You need ...

  10. POJ3468 A Simple Problem with Integers —— 线段树 区间修改

    题目链接:https://vjudge.net/problem/POJ-3468 You have N integers, A1, A2, ... , AN. You need to deal wit ...

随机推荐

  1. uva 1393 - Highways(容斥原理)

    题目连接:uva 1393 - Highways 题目大意:给定一个m∗n的矩阵,将矩阵上的点两两相连,问有多少条直线至少经过两点. 解题思路:头一次做这样的题目,卡了一晚上. dp[i][j]即为i ...

  2. Light OJ 1406 Assassin`s Creed 减少国家DP+支撑点甚至通缩+最小路径覆盖

    标题来源:problem=1406">Light OJ 1406 Assassin`s Creed 意甲冠军:向图 派出最少的人经过全部的城市 而且每一个人不能走别人走过的地方 思路: ...

  3. POJ 3422 Kaka&#39;s Matrix Travels(费用流)

    POJ 3422 Kaka's Matrix Travels 题目链接 题意:一个矩阵.从左上角往右下角走k趟,每次走过数字就变成0,而且获得这个数字,要求走完之后,所获得数字之和最大 思路:有点类似 ...

  4. 《数据结构、算法及应用》9.(C++实施订单)

    最近阅读<数据结构.算法及应用>这本书,书中的习题汇总,用自己的方法来实现这些问题.可能效率.等方面存在着非常多的问题,也可能是错误的实现.假设大家在看这本书的时候有更优更好的方法来实现, ...

  5. 发展,需求驱动 &#183; 一间 所见即所得

    从需求不是一句空话.同样是在发展过程中真正的. 需求驱动,与极限编程的一些想法和测试驱动开发基本重合. 鉴于该网站的发展是一个比较流行的方向,我会从网站开始,阐述自己的"需求驱动的发展&qu ...

  6. Windows 8实例教程系列 - 布局控制

    原文:Windows 8实例教程系列 - 布局控制 与传统应用类似,Windows store应用允许开发人员通过布局控件管理应用UI. 本篇将讨论Windows8布局设计控制. Windows 8布 ...

  7. JAVA: httpclient 详细说明——第四章;

    httpclient 具体解释--第一章. httpclient 具体解释--第二章: httpclient 具体解释--第三章: httpclient 具体解释--第四章: httpclient 具 ...

  8. asp.net webapi 多文件上传

    使用enctype="multipart/form-data"来进行操作 /// <summary> /// 上传图片 /// </summary> /// ...

  9. MyBatis与Spring设置callSettersOnNulls

    项目中集成Mybatis与Spring,使用的是Mybatis3.2.7,以及Spring4.0.5,mybatis-spring-1.2.2;由于项目组成员想要偷懒,将数据从DB中查询出来时须要将字 ...

  10. 使用Inno Setup 打包jdk、mysql、tomcat、webapp等为一个exe安装包(转)

    之前一直都没涉及到打包安装方面的东西,都是另一个同事负责的,使用的工具(installshield)也比较高大上一点,可是后来他离职以后接受的同事也只能是在这个基础上做个简单的配置,然后打包,可是现在 ...