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| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 92522 | Accepted: 28778 | |
| 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
1 #include <cstdio> 2 3 struct Node{ 4 int l, r; 5 long long val, add; 6 }; 7 Node node[100005<<2]; 8 9 void build(int v, int l, int r) 10 { 11 node[v].l = l; 12 node[v].r = r; 13 node[v].add = 0; 14 if(node[v].l == node[v].r){ 15 scanf("%I64d", &node[v].val); 16 return; 17 } 18 int mid = (l+r)>>1; 19 build(v<<1, l, mid); 20 build(v<<1|1, mid+1, r); 21 node[v].val = node[v<<1].val+node[v<<1|1].val; 22 } 23 24 void pushDown(int v) 25 { 26 if(node[v].add){ 27 node[v<<1].add += node[v].add; 28 node[v<<1|1].add += node[v].add; 29 int m = node[v].r-node[v].l+1; 30 node[v<<1].val += (m-(m>>1))*node[v].add; 31 node[v<<1|1].val += (m>>1)*node[v].add; 32 node[v].add = 0; 33 } 34 } 35 36 long long query(int v, int l, int r) 37 { 38 if(node[v].l == l && node[v].r == r) 39 return node[v].val; 40 pushDown(v); //更新子区间 41 int mid = (node[v].l+node[v].r)>>1; 42 if(r <= mid) 43 return query(v<<1, l, r); 44 else if(l > mid) 45 return query(v<<1|1, l, r); 46 else 47 return query(v<<1, l, mid)+query(v<<1|1, mid+1, r); 48 } 49 50 void update(int v, int l, int r, int c) 51 { 52 if(node[v].l == l && node[v].r == r){ 53 node[v].val += (r-l+1)*c; 54 node[v].add += c; 55 return; 56 } 57 pushDown(v); //更新子区间 58 int mid = (node[v].l+node[v].r)>>1; 59 if(r <= mid) 60 update(v<<1, l, r, c); 61 else if(l > mid) 62 update(v<<1|1, l, r, c); 63 else{ 64 update(v<<1, l, mid, c); 65 update(v<<1|1, mid+1, r, c); 66 } 67 node[v].val = node[v<<1].val+node[v<<1|1].val; 68 } 69 70 int main() 71 { 72 int N, Q; 73 scanf("%d%d", &N, &Q); 74 build(1, 1, N); 75 char order[5]; 76 int a, b, c; 77 while(Q--){ 78 scanf("%s", order); 79 if(order[0] == ‘Q‘){ 80 scanf("%d%d", &a, &b); 81 printf("%I64d\n", query(1, a, b)); 82 } 83 else{ 84 scanf("%d%d%d", &a, &b, &c); 85 update(1, a, b, c); 86 } 87 } 88 return 0; 89 }
POJ 3468 A Simple Problem with Integers
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原文地址：http://www.cnblogs.com/inmoonlight/p/5674834.html
