POJ 3468 A Simple Problem with Integers（线段树水题）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 135904		Accepted: 42113
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

 

思路：线段树水题，建议手写线段树，熟悉模板，注意数据需要使用long long

》》点击进入原题测试《《

#include<string>
#include<iostream>
#include<algorithm>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll long long

using namespace std;

const int maxn = 1e5 + 5;

struct Tree{
    ll l, r, w, f;
}tree[maxn << 2];
ll ans;

inline void PushDown(int rt)
{
    tree[rt << 1].f += tree[rt].f;
    tree[rt << 1 | 1].f += tree[rt].f;
    tree[rt << 1].w += tree[rt].f*(tree[rt << 1].r - tree[rt << 1].l + 1);
    tree[rt << 1 | 1].w += tree[rt].f*(tree[rt << 1 | 1].r - tree[rt << 1 | 1].l + 1);
    tree[rt].f = 0;
}
void build(int l, int r, int rt)
{
    tree[rt].l = l; 
    tree[rt].r = r;
    tree[rt].f = 0;
    if (l == r){
        cin >> tree[rt].w;
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    tree[rt].w = tree[rt << 1].w + tree[rt << 1 | 1].w;
}
void update(int L, int R, int l, int r, int rt)
{
    if (L <= l&& r <= R){
        tree[rt].w += ans*(r - l + 1);
        tree[rt].f += ans;
        return;
    }
    if (tree[rt].f)PushDown(rt);
    ll m = (l + r) >> 1;
    if (L <= m)update(L, R, lson);
    if (R > m) update(L, R, rson);
    tree[rt].w = tree[rt << 1].w + tree[rt << 1 | 1].w;
}
ll query(ll L, ll R, ll l, ll r, ll rt)
{
    if (L <= l&&r <= R){
        return tree[rt].w;
    }
    if (tree[rt].f)PushDown(rt);
    ll m = (l + r) >> 1, cnt = 0;
    if (L <= m)cnt +=  query(L, R, lson);
    if (R > m)cnt += query(L, R, rson);
    return cnt;
}
void Print(int l, int r, int rt)
{
    if (l == r){
        cout << rt << " = " << tree[rt].w << endl;
        return;
    }
    //cout << rt << " = " << tree[rt].w << endl;
    int m = (l + r) >> 1;
    if (l <= m)Print(lson);
    if (r > m)Print(rson);
}
int main()
{
    std::ios::sync_with_stdio(false);
    
    ll n, q; 
    cin >> n >> q;
        
    build(1, n, 1);
    string flag; ll a, b;
    while (q--){
        cin >> flag >> a >> b;
        if (flag == "C"){
            cin >> ans;;
            update(a, b, 1, n, 1);
            //Print(1, n, 1);
        }
        else if (flag == "Q"){
            cout << query(a, b, 1, n, 1) << endl;
        }
    }
    

    return 0;
}