POJ 3468 A Simple Problem with Integers (线段树)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 72740		Accepted: 22453
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

4
55
9
15



模板题。

#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using    namespace    std;

const    int    SIZE = 100005;
long    long    TREE[SIZE * 4],LAZY[SIZE * 4],N,Q;

void    build(int,int,int);
void    update(int,int,int,int,int,int);
void    push_down(int,int,int);
long    long    que(int,int,int,int,int);
int    main(void)
{
    char    op;
    int    a,b,c;

    while(~scanf("%d%d",&N,&Q))
    {
        build(1,1,N);
        while(Q --)
        {
            scanf(" %c%d%d",&op,&a,&b);
            if(op == 'Q')
                printf("%lld
",que(a,b,1,1,N));
            else
            {
                scanf("%d",&c);
                update(a,b,1,1,N,c);
            }
        }
    }

    return    0;
}

void    build(int node,int left,int right)
{
    LAZY[node] = 0;
    if(left == right)
        scanf("%lld",&TREE[node]);
    else
    {
        int    mid = (left + right) >> 1;
        build(node * 2,left,mid);
        build(node * 2 + 1,mid + 1,right);
        TREE[node] = TREE[node * 2] + TREE[node * 2 + 1];
    }
}

void    update(int L,int R,int node,int left,int right,int add)
{
    if(left >= L && right <= R)
    {
        TREE[node] += add * (right - left + 1);
        LAZY[node] += add ;
        return    ;
    }
    if(right < L || left > R)
        return    ;

    push_down(node,left,right);
    int    mid = (left + right) >> 1;
    update(L,R,node * 2,left,mid,add);
    update(L,R,node * 2 + 1,mid + 1,right,add);
    TREE[node] = TREE[node * 2] + TREE[node * 2 + 1];
}

long    long    que(int L,int R,int node,int left,int right)
{
    if(left >= L && right <= R)
        return    TREE[node];
    if(right < L || left > R)
        return    0;

    push_down(node,left,right);
    int    mid = (left + right) >> 1;
    return    que(L,R,node * 2,left,mid) + que(L,R,node * 2 + 1,mid + 1,right);
}

void    push_down(int node,int left,int right)
{
    if(LAZY[node])
    {
        int    mid = (left + right) >> 1;
        TREE[node * 2] += LAZY[node] * (mid - left + 1);
        LAZY[node * 2] += LAZY[node];
        TREE[node * 2 + 1] += LAZY[node] * (right - mid);
        LAZY[node * 2 + 1] += LAZY[node];
        LAZY[node] = 0;
    }
}