A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 53749 | Accepted: 16131 | |
| 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
//更新某一段区域的时候，采用延迟标记~~
代码：
1 #include "cstdio" //poj 3468 lazy操作 2 #include "cstring" 3 #include "iostream" 4 using namespace std; 5 6 #define N 100005 7 #define LL long long 8 9 struct node{ 10 int x,y; 11 LL sum; 12 LL add; //记录以当前节点为根节点的树中需要增加的值 13 }a[3*N]; 14 15 void Build(int t,int x,int y) 16 { 17 a[t].x = x; 18 a[t].y = y; 19 a[t].sum = a[t].add = 0; 20 if(a[t].x == a[t].y) //到了叶子节点 21 { 22 scanf("%lld",&a[t].sum); 23 return ; 24 } 25 int mid = (a[t].x + a[t].y)/2; 26 Build(t<<1,x,mid); 27 Build(t<<1|1,mid+1,y); 28 a[t].sum = a[t<<1].sum + a[t<<1|1].sum; 29 } 30 31 void Push_down(int t) //将add(增值)向下推一级 32 { 33 LL add = a[t].add; 34 a[t<<1].add += add; 35 a[t<<1|1].add += add; 36 a[t<<1].sum += add*(a[t<<1].y-a[t<<1].x+1); 37 a[t<<1|1].sum += add*(a[t<<1|1].y-a[t<<1|1].x+1); 38 a[t].add = 0; 39 } 40 41 LL Query(int t,int x,int y) 42 { 43 if(a[t].x==x &&a[t].y==y) 44 return a[t].sum; 45 Push_down(t); 46 int mid = (a[t].x + a[t].y)/2; 47 if(y<=mid) 48 return Query(t<<1,x,y); 49 if(x>mid) 50 return Query(t<<1|1,x,y); 51 else 52 return Query(t<<1,x,mid) + Query(t<<1|1,mid+1,y); 53 } 54 55 void Add(int t,int x,int y,int k) 56 { 57 if(a[t].x==x && a[t].y==y) //不在推到叶子节点，t下的子孙要增加的值存入a[t].add中 58 { 59 a[t].add += k; 60 a[t].sum += (a[t].y-a[t].x+1)*k; 61 return ; 62 } 63 a[t].sum += (y-x+1)*k; 64 Push_down(t); 65 int mid = (a[t].x+a[t].y)/2; 66 if(y<=mid) 67 Add(t<<1,x,y,k); 68 else if(x>mid) 69 Add(t<<1|1,x,y,k); 70 else 71 { 72 Add(t<<1,x,mid,k); 73 Add(t<<1|1,mid+1,y,k); 74 } 75 } 76 77 int main() 78 { 79 int n,m; 80 char ch; 81 int x,y,k; 82 scanf("%d %d",&n,&m); 83 Build(1,1,n); 84 while(m--) 85 { 86 getchar(); 87 scanf("%c %d %d",&ch,&x,&y); 88 if(ch=='Q') 89 printf("%lld ",Query(1,x,y)); 90 else 91 { 92 scanf("%d",&k); 93 Add(1,x,y,k); 94 } 95 } 96 return 0; 97 }