A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 31956		Accepted: 9089
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 51 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

, Yang Yi

//线段树进阶、成段更新

 

#include <iostream>

#include <stdio.h>

#include <string.h>

#define lson l,m,k<<1

#define rson m+1,r,k<<1|1

#define N 100001

using namespace std;

struct node

{

    long long sum,add;

};

node  st[N<<2];

void up(int &k)

{

    st[k].sum=st[k<<1].sum+st[k<<1|1].sum;

}

void down(int &k,int m)//将附加信心传递给孩子

{

    st[k<<1].add+=st[k].add;

    st[k<<1|1].add+=st[k].add;

    st[k<<1].sum+=st[k].add*(m-(m>>1));//这里注意m>>1要加括号、估计是优先级比较低呀

    st[k<<1|1].sum+=st[k].add*(m>>1);

    st[k].add=0;

}

void build(int l,int r,int k)

{

    st[k].add=0;

    if(l==r)

    {

        scanf("%lld",&st[k].sum);

        return ;

    }

    int m=(l+r)>>1;

    build(lson);

    build(rson);

    up(k);

}

long long add;

void updata(int &L,int &R,int l,int r,int k)

{

    if(L<=l&&R>=r)

    {

        st[k].add+=add;

        st[k].sum+=add*(r-l+1);//这里要注意，开始s[k].sum=s[k].add*（r-l+1）,当k是根时就会错的.

        return ;

    }

    if(st[k].add)

      down(k,r-l+1);

    int m=(l+r)>>1;

    if(L<=m) updata(L,R,lson);

    if(R>m)  updata(L,R,rson);

    up(k);

}

long long query(int &L,int &R,int l,int r,int k)

{

    if(L<=l&&R>=r)

    {

        return st[k].sum;

    }

    if(st[k].add)

        down(k,r-l+1);

    int m=(l+r)>>1;

    long long t1=0,t2=0;

    if(L<=m) t1=query(L,R,lson);

    if(R>m)  t2=query(L,R,rson);

    up(k);

    return t1+t2;

}

int main()

{

    int n,p;

    char c;

    while(scanf("%d%d",&n,&p)!=EOF)

    {

        build(1,n,1);

        int L,R;

        while(p--)

        {

            getchar();

          scanf("%c",&c);

          if(c=='Q')

          {

             scanf("%d%d",&L,&R);

             printf("%lld\n",query(L,R,1,n,1));

          }

          else

          {

             scanf("%d%d%lld",&L,&R,&add);

             updata(L,R,1,n,1);

          }

        }

    }

    return 0;

}