poj 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions:40260		Accepted: 11693
Case Time Limit: 2000MS

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 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

Hint

The sums may exceed the range of 32-bit integers.

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 #define lson l, m, rt<<1
 5 #define rson m+1, r, rt<<1|1
 6 #define LL long long int
 7 const int maxn = 111111;
 8 LL col[maxn<<2], sum[maxn<<2];
 9 void PushUP(int rt){
10   sum[rt] = sum[rt<<1] + sum[rt<<1|1];
11 }
12 void PushDown(int rt, int m){
13   if (col[rt]){
14     col[rt<<1] += col[rt]; col[rt<<1|1] += col[rt];
15     sum[rt<<1] += (LL)(col[rt] *(m-(m>>1)));
16     sum[rt<<1|1] += (LL)(col[rt] *(m>>1));
17     col[rt] = 0;
18   }
19 }
20 void build(int l, int r, int rt){
21   col[rt] = 0; if (l == r) {scanf("%lld", sum + rt); return;}
22   int m = (l + r) >> 1; build(lson); build(rson); PushUP(rt);
23 }
24 void update(int L, int R, int c, int l, int r, int rt){
25   if (L <= l && R >= r) {col[rt] += c; sum[rt] += (LL)(c*(r-l+1)); return;}
26   PushDown(rt, r - l + 1);
27   int m= (l + r) >> 1; 
28   if (L <= m) update(L, R, c, lson); if (R > m) update(L, R, c, rson);
29   PushUP(rt);
30 }
31 LL query(int L, int R, int l, int r, int rt){
32   if (L <= l && R >= r){return sum[rt];}
33   PushDown(rt, r - l + 1); int m = (l + r) >> 1; LL ret = 0;
34   if (L <= m) ret += query(L, R, lson); if (R > m) ret += query(L, R, rson);
35   PushUP(rt);
36   return ret;
37 }
38 int main(void){
39   int n, q;
40 #ifndef ONLINE_JUDGE
41   freopen("poj3468.in", "r", stdin);
42 #endif
43   while (~scanf("%d%d", &n, &q)){
44     build(1, n, 1);
45     while (q--){
46       char a[2]; scanf("%s", a); int m, b, c;
47       if (a[0] == 'Q') 
48       {scanf("%d%d", &m, &b);printf("%lld\n",query(m, b, 1, n, 1));}
49       else {
50         scanf("%d%d%d", &m, &b, &c); update(m, b, c, 1, n, 1);
51       }
52     }
53   }
54   return 0;
55 }

更新区间的线段树。敲了好几遍，差不多熟悉了，其实敲多了也就明白了。