poj 3468 A Simple Problem with Integers（线段树成段更新lazy）

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Description

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915
solution:
LAZY模板题~
#include<cstdio>#include<algorithm>using namespace std;const int maxn = 1e5+200;#define ls t<<1#define rs (t<<1)|1struct node{long long val,lazy;int l, r;}tree[maxn * 4];void pushup(int t){tree[t].val = tree[ls].val + tree[rs].val;}void build(int l, int r, int t){tree[t].l = l;tree[t].r = r;tree[t].lazy = 0;if (l == r){scanf("%lld", &tree[t].val);return;}build(l, (l + r) / 2, ls);build((l + r) / 2 + 1, r, rs);pushup(t);}void pushdown(int t){if (tree[t].lazy){tree[ls].lazy+= tree[t].lazy;tree[ls].val+= (tree[ls].r - tree[ls].l + 1)*tree[t].lazy;tree[rs].lazy+= tree[t].lazy;tree[rs].val+= (tree[rs].r - tree[rs].l + 1)*tree[t].lazy;tree[t].lazy = 0;}}void update(int l,int r,int t,long long k){if (tree[t].l >= l&&tree[t].r <= r){tree[t].lazy += k;tree[t].val+= (tree[t].r - tree[t].l + 1)*k;return;}pushdown(t);int mid = (tree[t].l + tree[t].r) / 2;if (l <= mid)update(l, r, ls, k);if (r > mid)update(l, r, rs, k);pushup(t);}long long Query(int l, int r, int t){long long sum = 0;if (tree[t].l >= l&&tree[t].r <= r)return tree[t].val;pushdown(t);int mid = (tree[t].l + tree[t].r) / 2;if (l <= mid)sum+=Query(l, r, ls);if (r > mid)sum+=Query(l, r, rs);return sum;}int main(){int t, n, m, nn, x, y;long long z;char op[10];while (~scanf("%d%d", &n, &m)){build(1, n, 1);while (m--){scanf("%s%d%d", op, &x, &y);if (op[0] == 'Q')printf("%lld\n", Query(x, y, 1));else {scanf("%lld", &z);update(x, y, 1, z);}}}}

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