A Simple Problem with Integers POJ
阿新 • • 發佈:2018-12-09
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
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; }