POJ 3468 A Simple Problem with Integers 线段树

A Simple Problem with Integers

Time Limit: 5000MS
Memory Limit: 131072KTotal Submissions: 56971
Accepted: 17284Case 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 andQ. 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

传送门：POJ 3468 A Simple Problem with Integers

题目分析：赤果果的区间修改区间查询，线段树搞之。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>typedef long long BigInt ;#define ll o << 1#define rr o << 1 | 1#define lson l , m , ll#define rson m + 1 , r , rrconst int maxN = 1000005 ;struct Seg_Tree {BigInt sum , add ;int l , r ;int len () {return r - l + 1 ;}int mid () {return ( l + r ) >> 1 ;}} tree[maxN] ;void PushUp ( int o ) {tree[o].sum = tree[ll].sum + tree[rr].sum ;}void PushDown ( int o ) {if ( tree[o].add ) {int len = tree[o].len () ;tree[ll].add += tree[o].add ;tree[rr].add += tree[o].add ;tree[ll].sum += tree[o].add * ( len - ( len >> 1 ) ) ;tree[rr].sum += tree[o].add * ( len >> 1 ) ;tree[o].add = 0 ;}}void Build ( int l , int r , int o ) {tree[o].l = l ;tree[o].r = r ;tree[o].add = 0 ;if ( tree[o].l == tree[o].r ) {scanf ( "%lld" , &tree[o].sum ) ;return ;}int m = tree[o].mid () ;Build ( lson ) ;Build ( rson ) ;PushUp ( o ) ;}void Update ( int L , int R , int o , int v ) {if ( L <= tree[o].l && tree[o].r <= R ) {tree[o].add += v ;tree[o].sum += v * tree[o].len () ;return ;}PushDown ( o ) ;int m = tree[o].mid () ;if ( L <= m ) Update ( L , R , ll , v ) ;if ( m <  R ) Update ( L , R , rr , v ) ;PushUp ( o ) ;}BigInt anssum ;void Query ( int L , int R , int o ) {if ( L <= tree[o].l && tree[o].r <= R ) {anssum += tree[o].sum ;return ;}PushDown ( o ) ;int m = tree[o].mid () ;if ( L <= m ) Query ( L , R , ll ) ;if ( m <  R ) Query ( L , R , rr ) ;}void work () {int n , m , L , R , v ;char ch[5] ;while ( ~scanf ( "%d%d" , &n , &m ) ) {Build ( 1 , n , 1 ) ;while ( m -- ) {scanf ( "%s" , ch ) ;if ( ch[0] == 'Q' ) {anssum = 0 ;scanf ( "%d%d" , &L , &R ) ;Query ( L , R , 1 ) ;printf ( "%lld\n" , anssum ) ;}if ( ch[0] == 'C' ) {scanf ( "%d%d%d" , &L , &R , &v ) ;Update ( L , R , 1 , v ) ;}}}}int main () {work () ;return 0 ;}