POJ3468(线段树区间维护)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 85502		Accepted: 26556
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

线段树入门题。

#include <cstdio>
using namespace std;
const int MAXN = 100005;
typedef long long LL;
struct Node{
    int l, r;
    LL sum, lazy;
}a[MAXN*3];
int n, m;
void build(int rt, int l, int r)
{
    a[rt].l = l;
    a[rt].r = r;
    a[rt].lazy = 0;
    if(l == r)
    {
        scanf("%I64d", &a[rt].sum);
        return ;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build((rt << 1) | 1, mid + 1, r);
    a[rt].sum = a[rt<<1].sum + a[(rt<<1)|1].sum;
}
void pushDown(int rt)
{
    int mid = (a[rt].l + a[rt].r) >> 1;
    a[rt<<1].sum += a[rt].lazy * (mid - a[rt].l + 1);
    a[(rt<<1)|1].sum += a[rt].lazy * (a[rt].r - mid);
    a[rt<<1].lazy += a[rt].lazy;
    a[(rt<<1)|1].lazy += a[rt].lazy;
    a[rt].lazy = 0;
}
void update(int rt, int l, int r, int val)
{
    if(a[rt].l == l && a[rt].r == r)
    {
        a[rt].sum += (LL)val * (r - l + 1);
        a[rt].lazy += (LL)val;
        return ;
    }
    if(a[rt].lazy != 0)
    {
        pushDown(rt);
    }
    int mid = (a[rt].l + a[rt].r) >> 1;
    if(r <= mid)
    {
        update(rt << 1, l, r, val);
    }
    else if(mid < l)
    {
        update((rt << 1) | 1, l, r, val);
    }
    else
    {
        update(rt << 1, l, mid, val);
        update((rt << 1) | 1, mid + 1, r, val);
    }
    a[rt].sum = a[rt<<1].sum + a[(rt<<1)|1].sum;
}
LL query(int rt, int l, int r)
{
    if(a[rt].l == l && a[rt].r == r)
    {
        return a[rt].sum;
    }
    if(a[rt].lazy != 0)
    {
        pushDown(rt);
    }
    int mid = (a[rt].l + a[rt].r) >> 1;
    if(r <= mid)
    {
        return query(rt << 1, l, r);
    }
    else if(mid < l)
    {
        return query((rt << 1) | 1, l, r);
    }
    else
    {
        return query(rt << 1, l, mid) + query((rt << 1) | 1, mid + 1, r);
    }
}
int main()
{
    while(scanf("%d %d",&n, &m) != EOF)
    {
        build(1, 1, n);
        while(m--)
        {
            scanf("%*c");
            char op;
            scanf("%c", &op);
            if(op == 'Q')
            {
                int l, r;
                scanf("%d %d", &l, &r);
                printf("%I64d
", query(1, l, r));
            }
            else
            {
                int l, r, val;
                scanf("%d %d %d", &l, &r ,&val);
                update(1, l, r, val);
            }
        }
    }
    return 0;
}