POJ 3468 区间更新，区间求和（经典）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 72265		Accepted: 22299
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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.

Source

POJ Monthly--2007.11.25, Yang Yi

 

题目意思：

给一个长度为n的数组，有q组操作，操作有两种:Q l, r  即查询l到r的和。  C l, r, val   即把l到r数组元素都加上val

 

思路：

线段树经典题目，需要用lazy思想，也就是当把l r加上val时，只标记这个区间lazy=val即可 ，当下次再加上一个val时且在l r之间，那么向下传递。

 

代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <set>
 using namespace std;

 #define N 100005
 #define ll root<<1
 #define rr root<<1|1
 #define mid (a[root].l+a[root].r)/2

 int max(int x,int y){return x>y?x:y;}
 int min(int x,int y){return x<y?x:y;}
 int abs(int x,int y){return x<?-x:x;}

 int n;

 struct node{
     int l, r;
     __int64 val, sum;
 }a[N*];

 void build(int l,int r,int root){
     a[root].l=l;
     a[root].r=r;
     a[root].val=;
     if(l==r){
         scanf("%I64d",&a[root].sum);
         return;
     }
     build(l,mid,ll);
     build(mid+,r,rr);
     a[root].sum=a[ll].sum+a[rr].sum;
 }

 void down(int root){
     if(a[root].val&&a[root].l!=a[root].r) {
         a[ll].val+=a[root].val;
         a[rr].val+=a[root].val;
         a[ll].sum+=(__int64)(a[ll].r-a[ll].l+)*a[root].val;
         a[rr].sum+=(__int64)(a[rr].r-a[rr].l+)*a[root].val;
         a[root].val=;
     }
 }

 void update(int l,int r,__int64 val,int root){
     if(a[root].l==l&&a[root].r==r){
         a[root].val+=val;
         a[root].sum+=(__int64)(a[root].r-a[root].l+)*val;
         return;
     }
     down(root);
     if(l>=a[rr].l) update(l,r,val,rr);
     else if(r<=a[ll].r) update(l,r,val,ll);
     else {
         update(l,mid,val,ll);
         update(mid+,r,val,rr);
     }
     a[root].sum=a[ll].sum+a[rr].sum;
 }

 __int64 query(int l,int r,int root){
     if(a[root].l==l&&a[root].r==r){
         return a[root].sum;
     }
     down(root);
     if(r<=a[ll].r) return query(l,r,ll);
     else if(l>=a[rr].l) return query(l,r,rr);
     else return query(l,mid,ll)+query(mid+,r,rr);
 }

 void out(int root){
     if(a[root].l==a[root].r) {
         printf("%I64d ",a[root].sum); return;
     }
     down(root);
     out(ll);
     out(rr);
 }
 main()
 {
     int i, j, k;
     int q;
     while(scanf("%d %d",&n,&q)==){
         build(,n,);
         char s[];
         __int64 w;
         int l, r;
         while(q--){
             scanf("%s",s);
             if(strcmp(s,"Q")==){
                 scanf("%d %d",&l,&r);
                 printf("%I64d\n",query(l,r,));
             }
             else{
                 scanf("%d %d %I64d",&l,&r,&w);
                 update(l,r,w,);
             //    out(1);
             }
         }
     }
 }