poj------(3468)A Simple Problem with Integers(区间更新)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 60745		Accepted: 18522
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

 

代码：

#include<cstdio>

#include<cstring>

const int maxn=100005;

struct node

{

 int lef,rig;

 __int64 sum,cnt;

 int mid(){

   return lef+(rig-lef>>1);

  }

};

 node reg[maxn<<2];



void Build(int left ,int right,int pos)

{

  reg[pos]=(node){left,right,0,0};

  if((left==right))

  {

    scanf("%I64d",&reg[pos].sum);

    return ;

  }

  int mid=reg[pos].mid();

  Build(left,mid,pos<<1);

  Build(mid+1,right,pos<<1|1);

  reg[pos].sum=reg[pos<<1].sum+reg[pos<<1|1].sum;

}

void Update(int left,int right,int pos,int val)

{

    if(reg[pos].lef>=left&&reg[pos].rig<=right)

    {

      reg[pos].cnt+=val;

      reg[pos].sum+=val*(reg[pos].rig-reg[pos].lef+1);

      return ;

    }

    if(reg[pos].cnt)

    {

      reg[pos<<1].cnt+=reg[pos].cnt;

      reg[pos<<1|1].cnt+=reg[pos].cnt;

      reg[pos<<1].sum+=reg[pos].cnt*(reg[pos<<1].rig-reg[pos<<1].lef+1);

      reg[pos<<1|1].sum+=reg[pos].cnt*(reg[pos<<1|1].rig-reg[pos<<1|1].lef+1);

      reg[pos].cnt=0;

    }

    int mid=reg[pos].mid();

    if(left<=mid)

        Update(left,right,pos<<1,val);

    if(right>mid)

        Update(left,right,pos<<1|1,val);

  reg[pos].sum=reg[pos<<1].sum+reg[pos<<1|1].sum;

}

__int64 Query(int left,int right,int pos)

{

    if(left<=reg[pos].lef&&reg[pos].rig<=right)

    {

      return reg[pos].sum;

    }

    if(reg[pos].cnt)  //再向下更新一次

    {

      reg[pos<<1].cnt+=reg[pos].cnt;

      reg[pos<<1|1].cnt+=reg[pos].cnt;

      reg[pos<<1].sum+=reg[pos].cnt*(reg[pos<<1].rig-reg[pos<<1].lef+1);

      reg[pos<<1|1].sum+=reg[pos].cnt*(reg[pos<<1|1].rig-reg[pos<<1|1].lef+1);

      reg[pos].cnt=0;

    }

    int mid=reg[pos].mid();

    __int64 res=0;

    if(left<=mid)

        res+=Query(left,right,pos<<1);

    if(mid<right)

        res+=Query(left,right,pos<<1|1);

   return res;

}

int main()

{

    int n,m,a,b,c;

    char ss;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        Build(1,n,1);

      while(m--)

       {

        getchar();

        scanf("%c %d%d",&ss,&a,&b);

        if(ss=='Q')

            printf("%I64d\n",Query(a,b,1));

         else{

           scanf("%d",&c);

           Update(a,b,1,c);

         }

       }

    }

  return 0;

}