【POJ 3468】A Simple Problem with Integers

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.

"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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.

区间更新，需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or 询问到的时候

线段树功能:update:成段增减 query:区间求和

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#define lson l, m, rt<<1
#define rson m + 1, r, rt<<1|1
using namespace std;
typedef long long ll;
const int MX = 1e6 + 5;
int n, q;
ll sum[MX << 2], add[MX << 2];
void PushUp(int rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void PushDown(int m, int rt)
{
if(add[rt])
{
sum[rt << 1] += add[rt] * (m - (m >> 1));
sum[rt << 1 | 1] += add[rt] * (m >> 1);
add[rt << 1] += add[rt];
add[rt << 1 | 1] += add[rt];
add[rt] = 0;
}
}
void build(int l, int r, int rt)
{
add[rt] = 0;
if(l == r)
{
scanf("%I64d", &sum[rt]);
return ;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void update(int L, int R, int c, int l, int r, int rt)
{
if(L <= l && r <= R)
{
sum[rt] += (ll)c * (r - l + 1);
add[rt] += c;
return ;
}
PushDown(r - l + 1, rt);
int m = (l + r) >> 1;
if(L <= m)
update(L, R, c, lson);
if(m < R)
update(L, R, c, rson);
PushUp(rt);
}
ll query(int L, int R, int l, int r, int rt)
{
if(L <= l && r <= R)
{
return sum[rt];
}
PushDown(r - l + 1, rt);
int m = (l + r) >> 1;
ll ret = 0;
if(L <= m)
ret += query(L, R, lson);
if(m < R)
ret += query(L, R, rson);
return ret;
}
int main()
{
scanf("%d%d", &n, &q);
build(1, n, 1);
for(int i = 0; i < q; i++)
{
int a, b, c;
char ch[2];
scanf("%s", &ch);
if(ch[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;
}