POJ 3468 A Simple Problem with Integers 成段更新，总区间求和

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 12302		Accepted: 2958
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

 

 

#include<stdio.h>

#define LL(x) (x<<1)

#define RR(x) (x<<1|1)

struct Seg_tree

{

    int left,right;

    __int64  sum,add;//用于标记下传，否则必TLE

    int clamid() {  return (left+right)/2; }

}tree[100001*3];

int num[100001];

int n,q;

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

{

    tree[idx].left=l;

    tree[idx].right=r;

    tree[idx].add=0;

    if(l==r)

    {

        tree[idx].sum=num[l];

        return ;

    }

    int mid=tree[idx].clamid();

    build(l,mid,LL(idx));

    build(mid+1,r,RR(idx));

    tree[idx].sum=tree[LL(idx)].sum+tree[RR(idx)].sum;

}

void Update(int l,int r,int idx,int c)

{

    if(l<=tree[idx].left&&r>=tree[idx].right)

    {

        tree[idx].add+=c;

        tree[idx].sum+=c*(tree[idx].right-tree[idx].left+1);

        return ;

    }

    if(tree[idx].add)

    {

        tree[LL(idx)].add+=tree[idx].add;

        tree[LL(idx)].sum+=tree[idx].add*(tree[LL(idx)].right-tree[LL(idx)].left+1);

        tree[RR(idx)].add+=tree[idx].add;

        tree[RR(idx)].sum+=tree[idx].add*(tree[RR(idx)].right-tree[RR(idx)].left+1);

        tree[idx].add=0;

    }

    int mid=tree[idx].clamid();

    if(r>mid) Update(l,r,RR(idx),c);

    if(l<=mid) Update(l,r,LL(idx),c);

    /*if(r>mid) Update(mid+1,r,RR(idx),c);

    if(l<=mid) Update(l,mid,LL(idx),c);这是错误的*/

    tree[idx].sum=tree[LL(idx)].sum+tree[RR(idx)].sum;

}

__int64  query(int l,int r,int idx)

{

    if(l<=tree[idx].left&&r>=tree[idx].right)

                 return tree[idx].sum;

    if(tree[idx].add)

    {

        tree[LL(idx)].add+=tree[idx].add;

        tree[LL(idx)].sum+=tree[idx].add*(tree[LL(idx)].right-tree[LL(idx)].left+1);

        tree[RR(idx)].add+=tree[idx].add;

        tree[RR(idx)].sum+=tree[idx].add*(tree[RR(idx)].right-tree[RR(idx)].left+1);

        tree[idx].add=0;

    }

    int mid=tree[idx].clamid();

    if(r<=mid) return query(l,r,LL(idx));

    else

    if(l>mid) return query(l,r,RR(idx));

    else

    return query(l,mid,LL(idx))+query(mid+1,r,RR(idx));

}

 

int main()

{

    while(scanf("%d%d",&n,&q)!=EOF)

    {

        for(int i=1;i<=n;i++)  scanf("%d",&num[i]);

        build(1,n,1);

        for(int i=1;i<=q;i++)

        {

            char str[10];

            int a,b,c;

            scanf("%s",str);

            if(str[0]=='Q')

            {

                scanf("%d%d",&a,&b);

                printf("%I64d/n",query(a,b,1));

            }

            else

            {

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

                Update(a,b,1,c);

            }

        }

    }

    return 0;

}