poj 3468 线段树对一个区间加减和求和

最新推荐文章于 2020-10-06 18:38:39 发布

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最新推荐文章于 2020-10-06 18:38:39 发布

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分类专栏：线段树文章标签： query numbers build rest each output

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本文介绍了一种处理整数序列的操作算法，包括在指定区间内加法运算和求和查询，支持超过32位整数范围的计算。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 10851		Accepted: 2550
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

#include <iostream> using namespace std; class cNode{ public:int l,r; long long value; long long delta;////对区间及子区间的改变值 }Node[400040]; int v[100010]; void build(int l,int r,int root) { if(l == r) { Node[root].value=v[l]; Node[root].delta=0; return; } { int mid=(l+r) / 2; build(l,mid,root*2); build(mid+1,r,root*2+1); Node[root].value=Node[root*2].value+Node[root*2+1].value; Node[root].delta=0; } } void Modify(int l,int r,int ll,int rr,int root,long long delta) { Node[root].value += (rr-ll+1)*delta; if(l == ll && r == rr) Node[root].delta+=delta; else { int mid=(l+r)/2; if(rr <= mid)//在左边 Modify(l,mid,ll,rr,2*root,delta); else if(ll >= mid+1)//在右边 Modify(mid+1,r,ll,rr,2*root+1,delta); else { Modify(l,mid,ll,mid,2*root,delta); Modify(mid+1,r,mid+1,rr,2*root+1,delta); } } } void Query(int l,int r,int ll,int rr,int root,__int64& cost) { if(l == ll && r == rr) { cost+=Node[root].value; return; } int mid=(l+r)/2; cost += (rr - ll +1)*Node[root].delta; if(rr <= mid) Query(l,mid,ll,rr,root*2,cost); else if(ll >= mid+1) Query(mid+1,r,ll,rr,root*2+1,cost); else { Query(l,mid,ll,mid,root*2,cost); Query(mid+1,r,mid+1,rr,root*2+1,cost); } } int main() { int n,order; while(scanf("%d%d",&n,&order)!=EOF) { //memset(Node,0,sizeof(Node)); memset(v,0,sizeof(v)); for(int i=1;i<=n;i++) scanf("%d",v+i); build(1,n,1); char oper; int a,b,c; for(int i=1;i<=order;i++) { cin>>oper; if(oper=='Q') { cin>>a>>b; __int64 rest=0; Query(1,n,a,b,1,rest); printf("%I64d/n",rest); } else { cin>>a>>b>>c; Modify(1,n,a,b,1,c); } } } }