A Simple Problem with Integers poj 3468 多树状数组解决区间修改问题。

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, ... , 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

 

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <sstream>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;

 LL sum[MS];
 LL C[][MS];
 int N,Q;

 int lowbit(int x)
 {
       return x&(-x);
 }

 void updata(int no,int x,LL value)
 {
       while(x<=N)
       {
             C[no][x]+=value;
             x+=lowbit(x);
       }
 }

 LL getsum(int no,int x)
 {
       LL res=;
       while(x>)
       {
             res+=C[no][x];
             x-=lowbit(x);
       }
       return res;
 }

 void solve()
 {
       scanf("%d%d",&N,&Q);
       sum[]=;
       for(int i=;i<=N;i++)
       {
             scanf("%lld",&sum[i]);
             sum[i]+=sum[i-];
       }
       memset(C,,sizeof(C));
       int l,r;
       LL c;
       char cmd[];
       for(int i=;i<=Q;i++)
       {
             scanf("%s",cmd);
             if(cmd[]=='Q')
             {
                   scanf("%d%d",&l,&r);
                   LL ans=sum[r]-sum[l-]+(getsum(,r)-getsum(,r)*(N-r))-(getsum(,l-)-getsum(,l-)*(N-l+));
                   printf("%lld\n",ans);
             }
             else
             {
                   scanf("%d%d%lld",&l,&r,&c);
                   updata(,l,c*(N-l+));
                   updata(,r+,(N-r)*(-c));

                   updata(,l,c);
                   updata(,r+,-c);
             }
       }
 }

 int main()
 {
       solve();
       return ;
 }