POJ-3468 A Simple Problem with Integers ( 线段树 )

题目链接: http://poj.org/problem?id=3468

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

4
55
9
15

询问区间和并对区间进行操作。
对区间操作是不能点更新，应该区间更新，不然会超时 例如区间(1,10)，我要对(3,8)进行操作，点更新需要进行(1,5),(1,3),(3,3),(4,5),(4,4),(5,5),(6,10),(6,8),(6,7),(6,6),(7,7),(8,8)共12步，而区间操作只需要(1,5),(1,3),(3,3),(4,5),(6,10),(6,8)共6步，而这只是一种简单的情况。
进行区间操作需要在节点加入一个参数add，用来记录该区间每个元素的增值以避免点更新。

#include<iostream>
#include<cstring>
using namespace std;

class node{
public:
    int l, r;
    long long v, add; //v为该区间的和，并用add记录加在该区间上的值
    node *left, *right;
};
long long num[100005];

node *Creat( int z, int y ){
    node *root = new node;
    root->l = z;
    root->r = y;
    root->add = 0;
    root->left = NULL;
    root->right = NULL;
    if( z == y ){
        root->v = num[z];
        return root;
    }

    int mid = ( z + y ) >> 1;
    root->left = Creat( z, mid );
    root->right = Creat( mid + 1, y );

    root->v = root->left->v + root->right->v;

    return root;
}

void Insert( int z, int y, long long n, node *p ){
    if( z == p->l && y == p->r ){
        p->add += n;
        return;
    }

    p->v += ( y - z + 1 ) * n;

    int mid = ( p->l + p->r ) >> 1;

    if( y <= mid ) Insert( z, y, n, p->left );
    else if( z > mid ) Insert( z, y, n, p->right );
    else{
        Insert( z, mid, n, p->left );
        Insert( mid +1, y, n, p->right );
    }
}

long long Ask( int z, int y,node *p ){
    long long ans = 0;

    if( z == p->l && y == p->r )
        return p->v + p->add * ( y - z +1 );

    ans += p->add * ( y - z + 1 );
    
    int mid = ( p->l + p->r ) >> 1;

    if( y <= mid ) ans += Ask( z, y, p->left );
    else if( z > mid ) ans += Ask( z, y, p->right );
    else
        ans += Ask( z, mid, p->left ) + Ask( mid + 1, y, p->right );
    
    return ans;
}

int main(){
    ios::sync_with_stdio( false );

    int N, Q;
    cin >> N >> Q;

    for( int i = 1; i <= N; i++ )
        cin >> num[i];

    node *root = Creat( 1, N );

    int a, b;
    long long c;
    char k;
    for( int i = 1; i <= Q; i++ ){
        cin >> k;
        
        if( k == 'C' ){
            cin >> a >> b >> c;
            Insert( a, b, c, root);
        }

        else{
            cin >> a >> b;
            cout << Ask( a, b, root ) << endl;
        }
    }

    return 0;
}