ACM学习历程——POJ3468 A Simple Problem with Integers（线段树）

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

Hint

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

 

这是对区间所有点增固定值类的线段树。

 

代码：

#include <iostream>

#include <cstdio>

#define LL long long



using namespace std;



int n, q;



//线段树

const int maxn = 100000;

struct node

{

    int lt, rt;

    LL val, add;

}tree[4*maxn];



//建立线段树

void Build(int lt, int rt, int id)

{

    tree[id].lt = lt;

    tree[id].rt = rt;

    tree[id].val = 0;//每段的初值，根据题目要求

    tree[id].add = 0;

    if (lt == rt)

    {

        scanf("%I64d", &tree[id].val);

        //tree[id].add = ??;

        return;

    }

    int mid = (lt + rt) >> 1;

    Build(lt, mid, id << 1);

    Build(mid + 1, rt, id << 1 | 1);

    tree[id].val = tree[id<<1].val + tree[id<<1|1].val;

}



void PushDown(int id, int pls)

{

    tree[id<<1].add += tree[id].add;

    //tree[id<<1].val += (pls-(pls>>1))*tree[id].add;

    tree[id<<1].val += (tree[id<<1].rt-tree[id<<1].lt+1)*tree[id].add;

    tree[id<<1|1].add += tree[id].add;

    //tree[id<<1|1].val += (pls>>1)*tree[id].add;

    tree[id<<1|1].val += (tree[id<<1|1].rt-tree[id<<1|1].lt+1)*tree[id].add;

    tree[id].add = 0;

}



//增加区间内每个点固定的值

void Add(int lt, int rt, int id, int pls)

{

    if (lt <= tree[id].lt && rt >= tree[id].rt)

    {

        tree[id].add += pls;

        tree[id].val += pls * (tree[id].rt-tree[id].lt+1);

        return;

    }

    if (tree[id].add != 0)

    {

        PushDown(id, tree[id].rt-tree[id].lt+1);

    }

    int mid = (tree[id].lt + tree[id].rt) >> 1;

    if (lt <= mid)

        Add(lt, rt, id<<1, pls);

    if (rt > mid)

        Add(lt, rt, id<<1|1, pls);

   tree[id].val = tree[id<<1].val + tree[id<<1|1].val;

}



LL Query(int lt, int rt, int id)

{

    if (lt <= tree[id].lt && rt >= tree[id].rt)

        return tree[id].val;

    if (tree[id].add != 0)

    {

        PushDown(id, tree[id].rt-tree[id].lt+1);

    }

    int mid = (tree[id].lt + tree[id].rt) >> 1;

    LL ans = 0;

    if (lt <= mid)

        ans += Query(lt, rt, id<<1);

    if (rt > mid)

        ans += Query(lt, rt, id<<1|1);

    return ans;



}



int main()

{

    //freopen("in.txt", "r", stdin);

    char op;

    int a, b, k;

    while (scanf("%d%d", &n, &q) != EOF)

    {

        Build(1, n, 1);

        for (int i = 0; i < q; ++i)

        {

            getchar();

            op = getchar();

            getchar();

            scanf("%d%d", &a, &b);

            if (op == 'Q')

                printf("%I64d\n", Query(a, b, 1));

            else

            {

                scanf("%d", &k);

                Add(a, b, 1, k);

            }

        }

    }

    return 0;

}