C - A Simple Problem with Integers （线段树lazy区间增减）

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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
long long int sum[400002];
long long int add[400002];
void PushUp(int node)
{
sum[node] = sum[node*2] + sum[node * 2 + 1];
}
//add数组标记
void PushDown(int node, int m)
{
if(add[node])
{
add[node*2] += add[node];
add[node*2+1] += add[node];
sum[node*2] += add[node] *(m - (m / 2));
sum[node*2+1] += add[node] *(m / 2);
add[node] = 0;
}
}
void Build(int l, int r, int node)
{
add[node] = 0;
if(l==r)
{
scanf("%lld", &sum[node]);
return ;
}
int m = (l + r) / 2;
Build(l, m , node * 2);
Build(m+1, r, node * 2 + 1);
PushUp(node);
}
void UpData(int L, int R, int Add, int l, int r, int node)
{
if(L<=l&&r<=R)
{
add[node] += Add;
sum[node] += (long long int)Add * (r - l + 1);
return ;
}
PushDown(node, r - l + 1);
int m = (l + r) / 2;
if(L<=m)
UpData(L, R, Add, l, m, node * 2);
if(m<R)
UpData(L, R, Add, m+1, r, node * 2 + 1);
PushUp(node);
}
long long int Query(int L, int R, int l, int r, int node)
{
if(L<=l&&r<=R)
{
return sum[node];
}
/*^^^^^^^^^^^^^^*/
PushDown(node, r - l + 1);
/*^^^^^^^^^^^^^^*/
long long int ans = 0;
int m = (l + r) / 2;
if(L<=m)
ans += Query(L, R, l, m, node * 2);
if(R>m)
ans += Query(L, R, m+1, r, node * 2 + 1);
return ans;
}
int main()
{
int n, m, a, b, c;
char op;
scanf("%d %d", &n, &m);
Build(1, n, 1);
getchar();
while(m--)
{
scanf("%c", &op);
if(op=='Q')
{
scanf("%d %d", &a, &b);
printf("%lld\n", Query(a, b, 1, n, 1));
}
else  if(op=='C')
{
scanf("%d %d %d", &a, &b, &c);
UpData(a, b, c, 1, n, 1);
}
getchar();
}
return 0;
}