A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 53255 | Accepted: 15926 | |
| 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
思路：线段树，节点区间的和由sum和add组成，更新：当区间正好匹配时就直接node[k].add += add,return，否则把和加在sum里，递归更新子区间。查找时:如果区间匹配就直接返回node[k].sum + (node[k].right - node[k].left + 1) * node[k].add，否则将和加在sum中，add向子区间传递。
AC代码：
1 /************************************************************************* 2 > File Name: inter.cpp 3 > Author: wangzhili 4 > Mail: wangstdio.h@gmail.com 5 > Created Time: 2014/2/27 星期四 14:16:52 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<cstdio> 10 using namespace std; 11 #define MAX 100005 12 class TreeNode 13 { 14 public: 15 int left; 16 int right; 17 int mid; 18 long long int sum; 19 long long int add; 20 }; 21 TreeNode node[4*MAX]; 22 void BuildTree(int k, int l, int r) 23 { 24 node[k].left = l; 25 node[k].right = r; 26 node[k].mid = (l + r) >> 1; 27 node[k].sum = 0; 28 node[k].add = 0; 29 if(l == r) 30 return ; 31 int mid = (l + r) >> 1; 32 BuildTree(k << 1, l, mid); 33 BuildTree(k << 1|1, mid+1, r); 34 } 35 36 void UpdateTree(int k, int l, int r, long long int add) 37 { 38 if(node[k].left == l && node[k].right == r) 39 { 40 node[k].add += add; 41 return ; 42 } 43 node[k].sum += (r-l+1) * add; 44 if(node[k].mid < l) 45 UpdateTree(k << 1|1, l, r, add); 46 else if(node[k].mid >= r) 47 UpdateTree(k << 1, l, r, add); 48 else 49 { 50 UpdateTree(k << 1, l, node[k].mid, add); 51 UpdateTree(k << 1|1, node[k].mid + 1, r, add); 52 } 53 } 54 55 long long int GetSum(int k, int l, int r) 56 { 57 if(node[k].left == l && node[k].right == r) 58 return node[k].sum + (node[k].right - node[k].left + 1) * node[k].add; 59 node[k << 1].add += node[k].add; 60 node[k << 1|1].add += node[k].add; 61 node[k].sum += (node[k].right - node[k].left + 1) * node[k].add; 62 node[k].add = 0; 63 if(node[k].mid < l) 64 return GetSum(k << 1|1, l, r); 65 else if(node[k].mid >= r) 66 return GetSum(k << 1, l, r); 67 else 68 return GetSum(k << 1, l, node[k].mid) + GetSum(k << 1|1, node[k].mid + 1, r); 69 } 70 71 int main(int argc, char const *argv[]) 72 { 73 int n, i, m, l, r, num; 74 char str[2]; 75 freopen("in.c", "r", stdin); 76 while(~scanf("%d%d", &n, &m)) 77 { 78 BuildTree(1, 1, n); 79 for(i = 1; i <= n; i ++) 80 { 81 scanf("%d", &num); 82 UpdateTree(1, i, i, num); 83 } 84 for(i = 0; i < m; i ++) 85 { 86 scanf("%s", str); 87 if(str[0] == 'Q') 88 { 89 scanf("%d%d", &l, &r); 90 printf("%lld ", GetSum(1, l, r)); 91 } 92 else 93 { 94 scanf("%d%d%d", &l, &r, &num); 95 UpdateTree(1, l, r, num); 96 } 97 } 98 } 99 return 0; 100 }