标签:eal str char log style cee 技术分享 memset show
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
The sums may exceed the range of 32-bit integers.
Means:
给你一段数列,有区间查询跟区间更新操作
Solve:
果Segment Tree || Binary Index Tree || Splay Tree
PS:线段树更新中有一段tree[ls] += (LL)lazy[id] * (mid - l + 1);这是由于更新的是区间,区间每个点都要加上value,所以要乘以区间点数
Code(SegmentTree):
1 #include <cstdio> 2 #include <cstring> 3 #define FOP freopen("in.in" , "r" , stdin) 4 #define FCL freopen("out.out" , "w" , stdout) 5 #define LL long long 6 #define MAXN 100000+10 7 #define CLR(x) memset(x , 0 , sizeof(x)) 8 #define LSON id << 1 , l , mid 9 #define RSON id << 1 | 1 , mid + 1 , r 10 #define ROOT 1 , 1 , n 11 #define PUSHUP(x) tree[id] = tree[id << 1] + tree[id << 1 | 1] 12 LL tree[MAXN << 2] = {0}; 13 LL data[MAXN] = {0}; 14 LL lazy[MAXN << 2] = {0}; 15 int n , q; 16 inline void PushDown(int id , int l , int r) 17 { 18 if(lazy[id]) 19 { 20 int ls = id << 1 , rs = id << 1 | 1 , mid = (l + r) >> 1; 21 lazy[ls] += lazy[id]; 22 lazy[rs] += lazy[id]; 23 tree[ls] += (LL)lazy[id] * (mid - l + 1); 24 tree[rs] += (LL)lazy[id] * (r - mid); 25 lazy[id] = 0; 26 } 27 } 28 void Build(int id , int l , int r) 29 { 30 if(l == r) 31 { 32 scanf("%lld" , tree + id); 33 lazy[id] = 0; 34 return ; 35 } 36 int mid = (l + r) >> 1; 37 Build(id << 1 , l , mid); 38 Build(id << 1 | 1 , mid + 1 , r); 39 PUSHUP(id); 40 } 41 LL Query(int id , int l , int r , int x , int y) 42 { 43 if(x <= l && y >= r) 44 { 45 return tree[id]; 46 } 47 PushDown(id , l , r); 48 int mid = (l + r) >> 1; 49 LL res = 0; 50 if(x <= mid) res += Query(LSON , x , y); 51 if(y > mid) res += Query(RSON , x , y); 52 PUSHUP(id); 53 return res; 54 } 55 void Update(int id , int l , int r , int x , int y , int z) 56 { 57 if(x <= l && y >= r) 58 { 59 lazy[id] += z; 60 tree[id] += (LL)(r - l + 1) * z; 61 return; 62 } 63 PushDown(id , l , r); 64 int mid = (l + r) >> 1; 65 if(x <= mid) Update(LSON , x , y , z); 66 if(y > mid) Update(RSON , x , y , z); 67 PUSHUP(id); 68 } 69 int main() 70 { 71 scanf("%d%d" , &n , &q); 72 Build(ROOT); 73 char cmd; 74 LL a , b , c; 75 while(q--) 76 { 77 scanf(" %c" , &cmd); 78 if(cmd == ‘Q‘) 79 { 80 scanf("%lld%lld" , &a , &b); 81 printf("%lld\n" , Query(ROOT , a , b)); 82 } 83 else 84 { 85 scanf("%lld%lld%lld" , &a , &b , &c); 86 Update(ROOT , a , b , c); 87 } 88 } 89 }
POJ3468-A Simple Problem with Integers(区间更新 + SegmentTree || SplayTree || BinaryIndexTree)
标签:eal str char log style cee 技术分享 memset show
原文地址:http://www.cnblogs.com/jianglingxin/p/6621812.html