A Simple Problem with Integers（树状数组，改段求段）

                               A Simple Problem with Integers

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXL=100000;
typedef long long int LL;
LL b[MAXL+50],c[MAXL+50];
LL sum[MAXL+50];
int a[MAXL+50];
int n,m;
int lowbit(int k)
{
    return k&-k;
}
void update(LL a[],int i,int value)
{
    while(i<=n)
    {
        a[i]+=value;
        i+=lowbit(i);
    }
}
LL getsum(LL a[],int i)
{
    LL ans=0;
    while(i>0)
    {
        ans+=a[i];
        i-=lowbit(i);
    }
    return ans;
}
int main()
{
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    memset(sum,0,sizeof(sum));
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++)
        scanf("%d",a+i);
    for(int i=1;i<=n;i++)
        sum[i]=sum[i-1]+a[i];
    int x,y,value;
    char ch;
    int T=m;
    while(T--)
    {
        getchar();
        scanf("%c",&ch);
        if(ch=='C')
        {
            scanf("%d%d%d",&x,&y,&value);
            update(b,x,value);
            update(b,y+1,-value);
            update(c,x,value*x);
            update(c,y+1,-value*(y+1));
        }
        else if(ch=='Q')
        {
            scanf("%d %d",&x,&y);
            LL ans=0;
            ans=sum[y]-sum[x-1];
            ans+=(y+1)*getsum(b,y)-getsum(c,y);
            ans-=(x*getsum(b,x-1))-getsum(c,x-1);
            cout<<ans<<endl;
        }

    }
}

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 100005
using namespace std;
typedef long long LL;
LL A[N], B[N], C[N];
int n, m;
void addB(int x, int k) //B[i]表示被1...i整体一共加了多少的总和
{
    for(int i=x; i<=n; i+=i&(-i))
        B[i]+=x*k;
}
void addC(int x, int k) //1....x节点的每个节点的增量
{
    for(int i=x; i>0; i-=i&(-i))
        C[i]+=k;
}
LL sumB(int x)
{
    LL s=0;
    for(int i=x; i>0; i-=i&(-i))
        s+=B[i];
    return s;
}
LL sumC(int x) //x节点总共的增量
{
    LL s=0;
    for(int i=x; i<=n; i+=i&(-i))
        s+=C[i];
    return s;
}
LL sum(int x)
{
    if(x==0)
        return 0;
    else
        return sumC(x)*x + sumB(x-1);
}
void update(int a, int b, int c)
{
    addB(b, c);
    addC(b, c);
    if(a-1>0)
    {
        addB(a-1, -c);
        addC(a-1, -c);
    }
}

int main()
{
    int m;
    while(scanf("%d %d", &n,&m)!=EOF)
    {
        for(int i=1; i<=n; ++i)
        {
            scanf("%lld", &A[i]);
            A[i]+=A[i-1];
        }
        int a, b, c;
        char u[2];
        while(m--)
        {
            scanf("%s ",u);
            if(u[0]=='Q')
            {
                scanf("%d%d", &a, &b);
                printf("%lld\n", A[b]-A[a-1]+sum(b)-sum(a-1));  //输出区间a到b的和

            }
            else
            {
                 scanf("%d%d%d", &a, &b, &c);
                 update(a, b, c);       //区间a到b加c
            }
        }
    }
    return 0;
}