POJ 3468 A Simple Problem with Integers (线段树区域更新)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 62431		Accepted: 19141
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

 

线段树的基础区域更新

#include<cstdio>

#include<iostream>

#include<cstring>

#include<stack>

#include<queue>

#include<cmath>

#include<stdlib.h>

#include<algorithm>

#define LL __int64

using namespace std;

const int MAXN=100000+5;

const int INF=0x3f3f3f3f;

struct node

{

    int l,r;

    LL num,flag;

    int mid()

    {

        return (l+r)>>1;

    }

}a[MAXN*4];

int b[MAXN],n,m;



void pushdown(int len,int step)

{

    if(a[step].flag)

    {

        a[step*2].flag+=a[step].flag;

        a[step*2+1].flag+=a[step].flag;

        a[step*2].num+=(len-len/2)*a[step].flag;

        a[step*2+1].num+=(len/2)*a[step].flag;

        a[step].flag=0;

    }

}



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

{

    a[step].l=l;

    a[step].r=r;

    a[step].flag=0;

    if(l==r)

    {

        a[step].num=b[l];

        return ;

    }

    int mid=a[step].mid();

    build(l,mid,step*2);

    build(mid+1,r,step*2+1);

    a[step].num=a[step*2].num+a[step*2+1].num;

}



void update(int x,int y,LL val,int step)

{

    if(x<=a[step].l && a[step].r<=y)

    {

        a[step].flag+=val;

        a[step].num+=(a[step].r-a[step].l+1)*val;

        return ;

    }

    pushdown(a[step].r-a[step].l+1,step);

    int mid=a[step].mid();

    if(x>mid) update(x,y,val,step*2+1);

    else if(y<=mid) update(x,y,val,step*2);

    else

    {

        update(x,mid,val,step*2);

        update(mid+1,y,val,step*2+1);

    }

    a[step].num=a[step*2].num+a[step*2+1].num;

}



LL query(int x,int y,int step)

{

    if(x<=a[step].l && a[step].r<=y) return a[step].num;

    pushdown(a[step].r-a[step].l+1,step);

    int mid=a[step].mid();

    if(x>mid) return query(x,y,step*2+1);

    else if(y<=mid) return query(x,y,step*2);

    else return query(x,y,step*2)+query(x,y,step*2+1);

}



int main()

{

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

    {

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

        build(1,n,1);

        while(m--)

        {

            char str[10];

            int A,B;

            scanf("%s %d %d",str,&A,&B);

            if(str[0]=='Q')

                printf("%I64d\n",query(A,B,1));

            if(str[0]=='C')

            {

                int cnt;

                scanf("%d",&cnt);

                update(A,B,cnt,1);

            }

        }

    }

    return 0;

}

View Code