A simple problem with integers POJ 3,468 multi-tree array solves the problem of interval modification.

A simple problem with integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 69589		Accepted: 21437
Case Time Limit: 2000MS

Description

You have N integers, a1, a2, ..., an. You need to deal with both kinds of operations. One type of operation is to add some given number to each number in a given interval. The other are to ask for the sum of numbers in a given interval.

Input

The first line contains the 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

2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ monthly--2007.11.25, Yang Yi

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <string>7#include <vector>8#include <Set>9#include <map>Ten#include <stack> One#include <queue> A#include <sstream> -#include <iomanip> - using namespacestd; thetypedefLong LongLL; - Const intinf=0x4fffffff; - Const intexp=1e-5; - Const intms=100005; +  - LL Sum[ms]; +LL c[2][ms]; A intn,q; at  - intLowbit (intx) - { -       returnx& (-x); - } -  in voidUpdata (intNointx,ll value) - { to        while(x<=N) +       { -c[no][x]+=value; thex+=lowbit (x); *       } $ }Panax Notoginseng  -LL Getsum (intNointx) the { +LL res=0; A        while(x>0) the       { +res+=C[no][x]; -x-=lowbit (x); $       } $       returnRes; - } -  the voidSolve () - {Wuyiscanf"%d%d",&n,&Q); thesum[0]=0; -        for(intI=1; i<=n;i++) Wu       { -scanf"%lld",&sum[i]); Aboutsum[i]+=sum[i-1]; $       } -memset (C,0,sizeof(C)); -       intL,r; - LL C; A       Charcmd[Ten]; +        for(intI=1; i<=q;i++) the       { -scanf"%s", cmd); $             if(cmd[0]=='Q') the             { thescanf"%d%d",&l,&R); theLL ans=sum[r]-sum[l-1]+ (Getsum (0, R)-getsum (1, R) * (N-R))-(Getsum (0, L-1)-getsum (1, L-1) * (n-l+1)); theprintf"%lld\n", ans); -             } in             Else the             { thescanf"%d%d%lld",&l,&r,&c); AboutUpdata (0, l,c* (n-l+1)); theUpdata (0, r+1, (N-R) * (-c)); the  theUpdata (1, l,c); +Updata (1, r+1,-c); -             } the       }Bayi } the  the  - intMain () - { the solve (); the       return 0; the}

A simple problem with integers POJ 3,468 multi-tree array solves the problem of interval modification.