Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 106215 | Accepted: 33148 | |
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
题意:C表示更新区间值,Q表示查询区间和,因为C和Q混杂,需要用线段树维护(若先C后Q,可以改用差分数列)。
# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define MAXN 100000 # define lson l, m, id<<1 # define rson m+1, r, id<<1|1 # define LL long long using namespace std; LL sum[MAXN<<2], lazy[MAXN<<2]; void build(int l, int r, int id) { lazy[id] = 0; if(l == r) { scanf("%lld",&sum[id]); return; } int m = (l+r)>>1; build(lson); build(rson); sum[id] = sum[id<<1] + sum[id<<1|1]; } void pushdown(int id, int len) { if(lazy[id]) { lazy[id<<1] += lazy[id]; lazy[id<<1|1] += lazy[id]; sum[id<<1] += lazy[id]*(len-(len>>1)); sum[id<<1|1] += lazy[id]*(len>>1); lazy[id] = 0; } } LL query(int L, int R, int l, int r, int id) { if(L <= l && R >= r) return sum[id]; pushdown(id, r-l+1);//不能放在上面。 LL ret = 0; int m = (l+r)>>1; if(L <= m) ret += query(L, R, lson); if(R > m) ret += query(L, R, rson); return ret; } void update(int L, int R, int k, int l, int r, int id) { if(L <= l && r <= R) { lazy[id] += k; sum[id] += (r-l+1)*k; return; } pushdown(id, r-l+1);//★因为下面要更新sum[id],所以要先更新下子区间。 int m = (l+r)>>1; if(L <= m) update(L, R, k, lson); if(R > m) update(L, R, k, rson); sum[id] = sum[id<<1] + sum[id<<1|1]; } int main() { int n, m, l, r, k; char c; while(~scanf("%d%d",&n,&m)) { build(1, n, 1); while(m--) { getchar(); c = getchar(); if(c == 'C') { scanf("%d%d%d",&l,&r,&k); update(l, r, k, 1, n, 1); } else { scanf("%d%d",&l,&r); printf("%lld\n",query(l, r, 1, n, 1)); } } } return 0; }