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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
//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 else52 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 else71 {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\n",Query(1,x,y));90 else91 {92 scanf("%d",&k);93 Add(1,x,y,k);94 }95 }96 return 0;97 }