POJ 3468 A Simple Problem with Integers （线段树【区间更新】）

最新推荐文章于 2019-05-04 02:44:36 发布

Anthony_azy

最新推荐文章于 2019-05-04 02:44:36 发布

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本文介绍了一个涉及区间操作的问题，包括区间内所有元素增加特定值及查询区间和的操作，并提供了一段使用线段树进行优化处理的C++代码实现。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 96409		Accepted: 30061
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.

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100000 + 10;
typedef long long ll;
int n, q;
char ch;
ll a, b, c;
ll s[maxn];
ll anssum;

struct node
{
    ll l, r;
    ll addv, sum;
}tree[maxn << 2];

void maintain(int id)
{
    if (tree[id].l >= tree[id].r)
        return;
    tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum;
}

void pushdown(int id)
{
    if (tree[id].l >= tree[id].r)
        return;
    if (tree[id].addv){
        int tmp = tree[id].addv;
        tree[id << 1].addv += tmp;
        tree[id << 1 | 1].addv += tmp;
        tree[id << 1].sum += (tree[id << 1].r - tree[id << 1].l + 1) * tmp;
        tree[id << 1 | 1].sum += (tree[id << 1 | 1].r - tree[id << 1 | 1].l + 1) * tmp;
        tree[id].addv = 0;
    }
}

void build(int id, ll l, ll r)
{
    tree[id].l = l;
    tree[id].r = r;
    tree[id].addv = 0;
    tree[id].sum = 0;
    if (l == r){
        tree[id].sum = s[l];
        return;
    }
    ll mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    maintain(id);
}

void updateAdd(int id, ll l, ll r, ll val)
{
    if (tree[id].l >= l && tree[id].r <= r){
        tree[id].addv += val;
        tree[id].sum += (tree[id].r - tree[id].l  + 1) * val;
        return;
    }
    pushdown(id);
    ll mid = (tree[id].l + tree[id].r) >> 1;
    if (l <= mid)
        updateAdd(id << 1, l, r, val);
    if (mid < r)
        updateAdd(id << 1 | 1, l, r, val);
    maintain(id);
}

void query(int id, ll l, ll r)
{
    if (tree[id].l >= l && tree[id].r <= r){
        anssum += tree[id].sum;
        return;
    }
    pushdown(id);
    ll mid = (tree[id].l + tree[id].r) >> 1;
    if (l <= mid)
        query(id << 1, l, r);
    if (mid < r)
        query(id << 1 | 1, l, r);
    maintain(id);
}

int main()
{
    while (scanf("%d%d", &n, &q) != EOF){
        for (int i = 1; i <= n; i++){
            scanf("%lld", &s[i]);
        }
        build(1, 1, n);
        while (q--){
            scanf("%*c%c", &ch);
            if (ch == 'C'){
                scanf("%lld%lld%lld", &a, &b, &c);
                updateAdd(1, a, b, c);
            }
            else{
                scanf("%lld%lld", &a, &b);
                anssum = 0;
                query(1, a, b);
                printf("%lld\n", anssum);
            }
        }
    }
    return 0;
}