(poj3468)A Simple Problem with Integers(区间更新)
来源:互联网 发布:全国淘宝店铺有多少个 编辑:程序博客网 时间:2024/05/22 10:37
Description
You have N integers, A1, A2, … , 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 A1, A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.
“Q a b” means querying the sum of Aa, Aa+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.
Source
POJ Monthly–2007.11.25, Yang Yi
分析:区间更新
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;const LL N=2e5+5;LL sum[N<<2],a[N],MAX[N<<2],ad[N<<2];void pushup(LL rt){ sum[rt]=sum[rt<<1]+sum[rt<<1|1]; // MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);}void pushdown(int rt,int m){ if(ad[rt]) { ad[rt<<1]+=ad[rt]; ad[rt<<1|1]+=ad[rt]; sum[rt<<1]+=ad[rt]*(m-(m>>1)); sum[rt<<1|1]+=ad[rt]*(m>>1); ad[rt]=0; ///标记清除!!! }}void build(LL l,LL r,LL rt){ ad[rt]=0; if(l==r) { scanf("%lld",&sum[rt]); return ; } LL m=(l+r)>>1; build(lson); build(rson); pushup(rt);}void update(LL L,LL R,LL c,LL l,LL r,LL rt){ if(L<=l&&R>=r) { ad[rt]+=c; sum[rt]+=c*(r-l+1); return ; } pushdown(rt,r-l+1); LL m=(l+r)>>1; if(L<=m) update(L,R,c,lson); if(R>m) update(L,R,c,rson); pushup(rt);}LL query(LL L,LL R,LL l,LL r,LL rt){ if(L<=l&&R>=r) { return sum[rt]; } pushdown(rt,r-l+1); LL m=(l+r)>>1; LL ret=0; if(L<=m) ret+=query(L,R,lson); if(R>m) ret+=query(L,R,rson); return ret;}int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { build(1,n,1); for(int i=0; i<m; i++) { char op; LL a,b,c; scanf(" %c",&op); if(op=='Q') { scanf("%lld%lld",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } else { scanf("%lld%lld%lld",&a,&b,&c); update(a,b,c,1,n,1); } } } return 0;}
- (poj3468)A Simple Problem with Integers(区间更新)
- POJ3468 A simple problem with integers(区间更新)
- POJ3468 A Simple Problem with Integers(区间更新)
- POJ3468 - A Simple Problem with Integers (线段树 区间更新)
- POJ3468 A Simple Problem with Integers(线段树 lazy思想 区间查询更新)
- poj3468——A Simple Problem with Integers(线段树,区间更新)
- POJ3468 A Simple Problem with Integers (线段树区间更新)
- poj3468 A Simple Problem with Integers(线段树+区间更新+非完全替换)模板
- poj3468 A Simple Problem with Integers 线段树区间更新
- 成段加值更新 区间求和 poj3468 A Simple Problem with Integers
- Poj3468 A Simple Problem with Integers 线段树、区间更新
- POJ3468 A Simple Problem with Integers 区间更新
- 【poj3468-A Simple Problem with Integers】-线段树区间更新
- poj3468 A Simple Problem with Integers(线段树成段更新)
- POJ3468--A Simple Problem with Integers(成段更新)
- poj3468 A Simple Problem with Integers(成段更新)
- POJ 3468 A Simple Problem with Integers(区间更新)
- POJ 3468 A Simple Problem with Integers(区间更新)
- Approximating a Constant Range CodeForces
- 动态规划:最长公共子序列
- IntelliJ IDEA 2017.2 破解(注册)
- 通用基础:序列化与反序列化
- cobbler常用命令及部署
- (poj3468)A Simple Problem with Integers(区间更新)
- PHP命名空间(Namespace)的使用详解
- 集成电路中低功耗设计(二)
- 远程协作与多方会议工具TeamViewer的使用
- iOS .a打进项目有的第三方库
- scala中的控制结构
- RedHat配置GPU计算环境
- MyBatis教程之五动态SQL的使用
- 如何让地址实现某一个长度的地址对齐?