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| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 81519 | Accepted: 25185 | |
| Case Time Limit: 2000MS | ||
Description
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
Source
/* * @Author: LinK * @Date: 2015-10-31 18:34:32 * @Last Modified by: LinK * @Last Modified time: 2015-10-31 23:11:18 */ #include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <vector> #include <cstdio> #include <string> #include <utility> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define randin srand((unsigned int)time(NULL)) #define input freopen("input.txt","r",stdin) #define debug(s) cout << "s = " << s << endl; #define outstars cout << "*************" << endl; const double PI = acos(-1.0); const int inf = 0x3f3f3f3f; const int INF = 0x7fffffff; typedef long long ll; const int maxn = 100010; struct Node { ll sum, add; } tree[maxn << 2]; int n, q; ll num[maxn]; void build(int rt, int l, int r) { if (l == r) { tree[rt].sum = num[l]; tree[rt].add = 0; return; } int mid = (l + r) >> 1; build(rt << 1, l, mid); build(rt << 1 | 1, mid + 1, r); tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; tree[rt].add = 0; } void update(int rt, int l, int r, int L, int R, ll val) { if (L <= l && R >= r) { tree[rt].add += val; return; } tree[rt].sum += val * (min(R, r) - max(L, l) + 1); int mid = (l + r) >> 1; if (R <= mid) update(rt << 1, l, mid, L, R, val); else if (L > mid) update(rt << 1 | 1, mid + 1, r, L, R, val); else { update(rt << 1, l, mid, L, R, val); update(rt << 1 | 1, mid + 1, r, L, R, val); } } ll query(int rt, int l, int r, int L, int R) { if (L <= l && R >= r) { return tree[rt].sum + tree[rt].add * (r - l + 1); } int mid = (l + r) >> 1; if (tree[rt].add) { tree[rt].sum += (r - l + 1) * tree[rt].add; tree[rt << 1].add += tree[rt].add; tree[rt << 1 | 1].add += tree[rt].add; tree[rt].add = 0; } if (R <= mid) return query(rt << 1, l, mid, L, R); if (L > mid) return query(rt << 1 | 1, mid + 1, r, L, R); return query(rt << 1, l, mid, L, R) + query(rt << 1 | 1, mid + 1, r, L, R); } int main() { char op; int a, b; ll c; while (~scanf("%d %d", &n, &q)) { for (int i = 1; i <= n; i ++) scanf("%lld", &num[i]); build(1, 1, n); while (q --) { scanf(" %c %d %d", &op, &a, &b); if (op == ‘Q‘) printf("%lld\n", query(1, 1, n, a, b)); else { scanf("%lld", &c); update(1, 1, n, a, b, c); } } } return 0; }
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原文地址：http://www.cnblogs.com/LinKArftc/p/4926523.html
