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A Simple Problem with Integers

Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 3468

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 abc" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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

 

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
const int maxx = 100010;
long long  tree[maxx<<2];
long long  addmark[maxx<<2];
long long a[maxx];
int n;
void build(int root,int l,int r)
{
    //cout<<"root:"<<root<<" l,r:"<<l<<"  "<<r<<endl;
    addmark[root]=0;
    if(l==r)
    {
        tree[root]=a[l];
    }
    else
    {
        int mid=(l+r)>>1;
        build(root<<1,l,mid);
        build((root<<1)+1,mid+1,r);
        tree[root]=(tree[root<<1]+tree[(root<<1)+1]);
    }
}
void push_down(int root,int l,int r)
{
    if(addmark[root]!=0)
    {
        int mid=(l+r)>>1;
        addmark[root<<1]+=addmark[root];
        addmark[(root<<1)+1]+=addmark[root];
        tree[root<<1]+=addmark[root]*(mid-l+1);
        tree[(root<<1)+1]+=addmark[root]*(r-(mid+1)+1);
        addmark[root]=0;
    }
}
void update(int root,int l,int r,int pos,int val)
{
    if(pos==l&&pos==r)
    {
        tree[root]+=val;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid) update(root<<1,l,mid,pos,val);
    else update((root<<1)+1,mid+1,r,pos,val);
    tree[root]=(tree[root<<1]+tree[(root<<1)+1]);
}

void update_interval(int root,int l,int r,int ql,int qr,int val)
{
    if(l==ql&&r==qr)
    {
        tree[root]+=val*(qr-ql+1);
        addmark[root]+=val;
        return;
    }
    push_down(root,l,r);
    int mid=(l+r)>>1;
    if(qr<=mid)
    {
        update_interval(root<<1,l,mid,ql,qr,val);
    }
    else if(ql>mid)
    {
        update_interval((root<<1)+1,mid+1,r,ql,qr,val);
    }
    else
    {
        update_interval(root<<1,l,mid,ql,mid,val);
        update_interval((root<<1)+1,mid+1,r,mid+1,qr,val);
    }
    tree[root]=tree[root<<1]+tree[(root<<1)+1];
}
long long  query(int root,int l,int r,int ql,int qr)
{

    if(l==ql&&qr==r)
    {
        return tree[root];
    }
    int mid=(l+r)>>1;
    if(addmark[root]!=0) push_down(root,l,r);
    if(qr<=mid)
    {
        return query(root<<1,l,mid,ql,qr);
    }
    else if(ql>mid)
    {
        return query((root<<1)+1,mid+1,r,ql,qr);
    }

    else return (query(root<<1,l,mid,ql,mid)+query((root<<1)+1,mid+1,r,mid+1,qr));
}

void print()
{
    int j=1,t=1;
    for(int i=1; i<=(n<<2); i++)
    {
        j++;
        printf("%I64d ",tree[i]);
        if(j>pow(2,(t-1)))
        {
            printf("
");
            j=1;
            t++;
        }

    }
    printf("

");
}
int main()
{
    int m;
    scanf("%d%d",&n,&m);

        for(int i=1; i<=n; i++)
        {
            scanf("%I64d",a+i);
        }
        build(1,1,n);

        char  cmd;
        for(int i=0; i<m; i++)
        {
            cin>>cmd;
            if(cmd=='C')
            {
                int tmp1,tmp2,val;
                scanf("%d%d%d",&tmp1,&tmp2,&val);
                update_interval(1,1,n,tmp1,tmp2,val);
            }
            else if(cmd=='Q')
            {
                int tmp1,tmp2;
                scanf("%d%d",&tmp1,&tmp2);
                long long ans=query(1,1,n,tmp1,tmp2);
                printf("%I64d
",ans);
            }
        }

    return 0;
}