POJ 3468 线段树，OIer杨弋出的题目。。膜拜中。。。orz...

最新推荐文章于 2018-10-05 17:00:56 发布

最新推荐文章于 2018-10-05 17:00:56 发布 · 106 阅读

本文介绍了一个涉及整数数组的操作问题，包括对指定区间内的整数进行加法操作及查询区间内整数之和。通过构建一种特殊的树形数据结构来高效处理这两种操作，确保即使面对大量请求也能快速响应。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 17755		Accepted: 4605
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<stdio.h> #include<string.h> struct node { __int64 l; __int64 r; __int64 d; __int64 val; }; node tree[400000]; __int64 num[100005],sum; void build(__int64 l,__int64 r,__int64 root) { tree[root].l=l; tree[root].r=r; if(l==r) { tree[root].val=num[l]; tree[root].d=0; return; } __int64 mid=(l+r)>>1; build(l,mid,root*2); build(mid+1,r,root*2+1); tree[root].val=tree[root*2].val+tree[root*2+1].val; tree[root].d=0; } void findsum(__int64 l,__int64 r,__int64 root) { if(tree[root].l==l&&tree[root].r==r) { sum=sum+tree[root].val; return; } sum=sum+(r-l+1)*tree[root].d; __int64 mid=(tree[root].l+tree[root].r)>>1; if(mid>=r) findsum(l,r,root*2); else if(mid<l) findsum(l,r,root*2+1); else { findsum(l,mid,root*2); findsum(mid+1,r,root*2+1); } } void update(__int64 l,__int64 r,__int64 c,__int64 root) { tree[root].val=tree[root].val+((r-l+1)*c); if(tree[root].l==l&&tree[root].r==r) { tree[root].d=tree[root].d+c; return; } __int64 mid=(tree[root].l+tree[root].r)>>1; if(mid>=r) update(l,r,c,root*2); else if(mid<l) update(l,r,c,root*2+1); else { update(l,mid,c,root*2); update(mid+1,r,c,root*2+1); } } int main() { __int64 i,n,Q,a,b,c; char ch[2]; scanf("%I64d%I64d",&n,&Q); for(i=1;i<=n;i++) scanf("%I64d",&num[i]); build(1,n,1); while(Q--) { scanf("%s",ch); if(ch[0]=='Q') { sum=0; scanf("%I64d%I64d",&a,&b); findsum(a,b,1); printf("%I64d/n",sum); } if(ch[0]=='C') { scanf("%I64d%I64d%I64d",&a,&b,&c); update(a,b,c,1); } } return 0; }