POJ - 3468-A Simple Problem with Integers (树状数组)

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.

树状数组写法:

#include<algorithm>
#include<string.h>
#include<stdio.h>
#define M 100010
using namespace std;
long long a[M],T[M],Ti[M];
int n,m;
long long lowbit(long long b)
{
    return b&-b;
}
void add(long long *arry,long long pos,long long x)
{
    while(pos<=n)
    {
        arry[pos]+=x;
        pos+=lowbit(pos);
    }
}
long long query(long long *arry,long long pos)
{
    long long ans=0;
    while(pos>0)
    {
        ans+=arry[pos];
        pos-=lowbit(pos);
    }
    return ans;
}
int main()
{
    long long l,r,x,sumr,suml;
    char str[10];
    scanf("%d%d",&n,&m);
    for(long long i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        a[i]+=a[i-1];
    }
    while(m--)
    {
        scanf("%s",str);
        if(str[0]=='C')
        {
            scanf("%lld%lld%lld",&l,&r,&x);
            add(T,l,x);
            add(T,r+1,-x);
            add(Ti,l,l*x);
            add(Ti,r+1,-x*(r+1));
        }
        else
        {
            scanf("%lld%lld",&l,&r);
            suml=a[l-1]+l*query(T,l)-query(Ti,l);
            sumr=a[r]+(r+1)*query(T,r)-query(Ti,r);
            printf("%lld\n",sumr-suml);
        }
    }
}