A Simple Problem with Integers

最新推荐文章于 2025-05-27 22:22:47 发布

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最新推荐文章于 2025-05-27 22:22:47 发布

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分类专栏：线段树

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本文介绍了一种处理数组中特定区间加法操作及求和查询的高效算法实现。通过使用线段树数据结构，该算法能在对数时间内完成更新与查询操作，适用于需要大量区间修改与查询的应用场景。

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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 abc" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
#define INF 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
const int maxn=100010;
int n,q;
long long sum[maxn<<2];
long long add[maxn<<2];

void Push_Up(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void Push_Down(int rt,int m)
{
    if (add[rt])
    {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt] * (m - (m >> 1));
        sum[rt<<1|1] += add[rt] * (m >> 1);
        add[rt] = 0;
    }
}

void build(int l,int r,int rt)
{
    add[rt] = 0;
    if (l == r)
    {
        scanf("%I64d",&sum[rt]);
        return ;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    Push_Up(rt);
}

void update(int L,int R,int c,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        add[rt] += c;
        sum[rt] += (long long)c * (r - l + 1);
        return ;
    }
    Push_Down(rt , r - l + 1);
    int mid = (l + r) >> 1;
    if (L <= mid) update(L , R , c , lson);
    if (mid < R) update(L , R , c , rson);
    Push_Up(rt);
}

long long query(int L,int R,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        return sum[rt];
    }
    Push_Down(rt , r - l + 1);
    int mid = (l + r) >> 1;
    long long ret = 0;
    if (L <= mid) ret += query(L , R , lson);
    if (mid < R) ret += query(L , R , rson);
    return ret;
}

int main()
{

    while(~scanf("%d%d",&n,&q))
    {
        build(1,n,1);
        char op[3];
        int a,b,c;
        while(q--)
        {
            scanf("%s",op);
            if (op[0] == 'Q')
            {
                scanf("%d%d",&a,&b);
                printf("%I64d\n",query(a , b , 1 , n , 1));
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a , b , c , 1 , n , 1);
            }
        }
    }
    return 0;
}