poj 3468 A Simple Problem with Integers 【线段树 + 区间更新lazy】

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最新推荐文章于 2016-07-05 10:41:57 发布

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分类专栏：线段树

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本文探讨了在处理大量整数时，如何高效地执行区间加法和区间求和操作，涉及复杂的数据结构和算法应用。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 71894		Accepted: 22174
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

Hint

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

ac:

#include<cstdio>
#include<cstring>
#define MAX 100000+10 
using namespace std;
long long add[MAX<<2];
long long sum[MAX<<2];
void PushUp(int o)
{
    sum[o] = sum[o<<1] + sum[o<<1|1];
}
void PushDown(int o, int m)
{
    if(add[o])
    {
        add[o<<1]   += add[o];
        add[o<<1|1] += add[o];
        sum[o<<1]   += add[o] * (m-(m>>1));
        sum[o<<1|1] += add[o] * (m>>1);
        add[o] = 0;
    }
}
void build(int o, int l, int r)
{
    add[o] = 0;
    if(l == r)
    {
        scanf("%lld", &sum[o]);
        return ;
    }
    int mid = (l + r) >> 1;
    build(o<<1, l, mid);
    build(o<<1|1, mid+1, r);
    PushUp(o);
}
void update(int o, int l, int r, int L, int R, int v)
{
    if(L <= l && R >= r)
    {
        add[o] += v;
        sum[o] += v * (r-l+1);
        return ;
    }
    PushDown(o,r-l+1);
    int mid = (r+l) >> 1;
    if(L <= mid)
    update(o<<1, l, mid, L ,R, v);
    if(R > mid)
    update(o<<1|1, mid+1, r, L ,R, v);
    PushUp(o);
}
long long query(int o, int l, int r, int L, int R)
{
    if(L <= l && R >= r)
    {
        return sum[o];
    }
    PushDown(o,r-l+1);
    int mid = (r+l) >> 1;
    long long res = 0;
    if(L <= mid)
    res += query(o<<1, l, mid, L ,R);
    if(R > mid)
    res += query(o<<1|1, mid+1, r, L, R);
    return res;
}
int main()
{
    int n, m;
    int a, b, v;
    char op[3];
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        build(1, 1, n);
        while(m--)
        {
            scanf("%s",op);
            if(op[0] == 'Q')
            {
                scanf("%d%d", &a, &b);
                printf("%lld\n",query(1, 1, n, a ,b));
            }
            else
            {
                scanf("%d%d%d", &a, &b, &v);
                update(1, 1, n ,a ,b ,v);
            }
        }
    }
    return 0;
}