解题报告

题意：

略

思路：

线段树成段更新，区间求和。

#include 
  
   
#include 
   
     #include 
    
      #define LL long long #define int_now int l,int r,int root using namespace std; LL sum[500000],lazy[500000]; void push_up(int root,int l,int r) { sum[root]=sum[root*2]+sum[root*2+1] + lazy[root]*(r-l+1); } void update(int root,int l,int r,int ql,int qr,LL v) { if(ql>r||qr
     
      r||qr
      
       
 
       
 A Simple Problem with Integers 
       
 
        
 
         
 
          
Time Limit: 5000MS
 
          
 
 
          
Memory Limit: 131072K
 
         
 
         
 
          
Total Submissions: 60817
 
          
 
 
          
Accepted: 18545
 
         
 
         
 
          
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
 
       
Hint
 The sums may exceed the range of 32-bit integers.