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 <algorithm>
#include <cstdio>
using namespace std;
const int N = 1e6 + 7;
struct Tree {
int l, r, lazy;
long long sum;
};
struct Tree tree[N];
long long a[N];
void pushup (int rt) {
//printf ("pushup!\n");
tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
}
void build (int l, int r, int rt) {
tree[rt].l = l, tree[rt].r = r, tree[rt].lazy = 0, tree[rt].sum = 0;
if (l == r) {
tree[rt].sum = a[l];
return;
}
int mid = (l + r) >> 1;
build (l, mid, rt << 1);
build (mid + 1, r, rt << 1 | 1);
pushup (rt);
}
void pushdown (int rt) {
//printf ("pushdown!\n");
int mid = (tree[rt].l + tree[rt].r) >> 1;
tree[rt << 1].sum += (long long)tree[rt].lazy * (mid - tree[rt].l + 1); // +=
tree[rt << 1 | 1].sum += (long long)tree[rt].lazy * (tree[rt].r - mid);
tree[rt << 1].lazy += tree[rt].lazy; // +=
tree[rt << 1 | 1].lazy += tree[rt].lazy;
tree[rt].lazy = 0;
}
void update (int l, int r, int val, int rt) {
//printf ("update!\n");
if (tree[rt].l == l && tree[rt].r == r) {
tree[rt].sum += (long long)(val * (r - l + 1));
tree[rt].lazy += val; // +=
return;
}
if (tree[rt].lazy != 0) pushdown (rt);
int mid = (tree[rt].l + tree[rt].r) >> 1;
if (r <= mid) update (l, r, val, rt << 1);
else if (l > mid) update (l, r, val, rt << 1 | 1);
else {
update (l, mid, val, rt << 1);
update (mid + 1, r, val, rt << 1 | 1);
}
pushup (rt);
}
long long query (int l, int r, int rt) {
//printf ("%d %d %d %d %lld\n", l, r, tree[rt].l, tree[rt].r, tree[rt].sum);
if (tree[rt].l == l && tree[rt].r == r) {
return tree[rt].sum;
}
if (tree[rt].lazy != 0) pushdown (rt);
int mid = (tree[rt].l + tree[rt].r) >> 1;
if (r <= mid) return query (l, r, rt << 1);
else if (l > mid) return query (l, r, rt << 1 | 1);
else return (long long)query (l, mid, rt << 1) + query (mid + 1, r, rt << 1 | 1);
}
int main () {
int n, q;
while (~scanf ("%d %d", &n, &q)) {
for (int i = 1; i <= n; ++i) {
scanf ("%lld", &a[i]);
}
build (1, n, 1);
char op;
int x, y, z;
while (q--) {
scanf (" %c %d %d", &op, &x, &y);
if (op == 'C') {
scanf ("%d", &z);
update (x, y, z, 1);
} else {
printf ("%lld\n", query (x, y, 1));
}
}
}
return 0;
}