A Simple Problem with Integers POJ - 3468 区间修改线段树lazy—tag大法

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

Hint

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

傻逼poj不支持万能头文件

//#include <bits/stdc++.h>
#include <stdio.h>
using namespace std;
typedef long long ll;
const ll inf = 4e18+10;
const int mod = 1000000007;
const int mx = 5e6+5; //check the limits, dummy
//typedef pair<int, int> pa;
//const double PI = acos(-1);
//ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
//#define swa(a,b) a^=b^=a^=b
//#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
//#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
//#define clr(a) memset(a, 0, sizeof(a))
//#define lowbit(x) ((x)&(x-1))
//#define mkp make_pair
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
ll sum[mx], add[mx];
void pushup(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void pushdown(int rt, int m) {//更新rt的子节点
    if (add[rt]) {
        add[rt << 1] += add[rt];
        add[rt << 1 | 1] += add[rt];
        sum[rt << 1] += (m - (m >> 1))* add[rt];
        sum[rt << 1 | 1] += (m >> 1)* add[rt];
        add[rt] = 0;//取消本层标记
    }
}
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
void biuld(int l, int r, int rt) {//用满二叉树建树
    add[rt] = 0;
    if (l == r) {//叶子结点，赋值
        scanf("%lld", &sum[rt]);
        return;
    }
    int mid = (l + r) >> 1;
    biuld(lson);
    biuld(rson);
    pushup(rt);//向上更新区间和
}
void update(int a, int b, ll c, int l, int r, int rt) {//区间更新
    if (a <= l && b >= r) {
        sum[rt] += (r - l + 1) * c;
        add[rt] += c;
        return;
    }
    pushdown(rt, r - l + 1);//向下更新
    int mid = (l + r) >> 1;
    if (a <= mid)update(a, b, c, lson);//分成两半深入
    if (b > mid)update(a, b, c, rson);
    pushup(rt);
}
ll query(int a, int b, int l, int r, int rt) {//区间求和
    if (a <= l && b >= r)
        return sum[rt];//满足lazy直接返回
    pushdown(rt, r - l + 1);
    int mid = (l + r) >> 1;
    ll ans = 0;
    if (a <= mid)ans += query(a, b, lson);
    if (b > mid)ans += query(a, b, rson);
    return ans;
}
int main()
{
    //ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int n, m;
    scanf("%d", &n); scanf("%d", &m);
    biuld(1, n, 1);
    while (m--) {
        char str[2];
        int a, b;
        ll c;
        scanf("%s", str);
        if (str[0] == 'C') {
            scanf("%d", &a); scanf("%d", &b); scanf("%lld", &c);
            update(a, b, c, 1, n, 1);
        }
        else {
            scanf("%d", &a); scanf("%d", &b);
            printf("%lld\n", query(a, b, 1, n, 1));
        }
    }
    return 0;
}