BZOJ-3212 Pku3468 A Simple Problem with Integers 裸线段树区间维护查询

3212: Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 1278 Solved: 560

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Description

You have N integers, A1, A2, … , AN. 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, … , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.

“Q a b” means querying the sum of Aa, Aa+1, … , Ab.

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.

题目大意：区间修改和区间查询

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代码如下：

/**************************************************************

    Problem: 3212

    User: DaD3zZ

    Language: C++

    Result: Accepted

    Time:48 ms

    Memory:8304 kb

****************************************************************/

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

#define maxn 100001

long long sum[maxn<<2]={0},delta[maxn<<2]={0};

long long a[maxn]={0};

void updata(int now)

{

    sum[now]=sum[now<<1]+sum[now<<1|1];

}

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

{

    if (l==r)

        {

            sum[now]=a[l];

            return;

        }

    int mid=(l+r)>>1;

    build(now<<1,l,mid);

    build(now<<1|1,mid+1,r);

    updata(now);

}

void pushdown(int now,int ln,int rn)

{

    if (delta[now]!=0)

        {

            delta[now<<1]+=delta[now];

            delta[now<<1|1]+=delta[now];

            sum[now<<1]+=delta[now]*ln;

            sum[now<<1|1]+=delta[now]*rn;

            delta[now]=0;

        }

}

void change(int L,int R,int now,int l,int r,long long data)

{

    if (L<=l && R>=r)

        {

            sum[now]+=data*(r-l+1);

            delta[now]+=data;

            return;

        }

    int mid=(l+r)>>1;

    pushdown(now,mid-l+1,r-mid);//这里需要先下放一波标记不然会出错

    if (L<=mid)

        change(L,R,now<<1,l,mid,data);

    if (R>mid)

        change(L,R,now<<1|1,mid+1,r,data);

    updata(now);

}

long long query(int L,int R,int now,int l,int r)

{

    if (L<=l && R>=r)

        return sum[now];

    int mid=(l+r)>>1;

    pushdown(now,mid-l+1,r-mid);

    long long total=0;

    if (L<=mid)

        total+=query(L,R,now<<1,l,mid);

    if (R>mid)

        total+=query(L,R,now<<1|1,mid+1,r);

    return total;

}

int main()

{

    int n,m;

    scanf("%d%d",&n,&m);

    for (int i=1; i<=n; i++)

        scanf("%lld",&a[i]);

    build(1,1,n);

    for (int i=1; i<=m; i++)

        {

            char c[3];

            int l,r;

            scanf("%s",&c);

            scanf("%d%d",&l,&r);

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

                {

                    long long ans=query(l,r,1,1,n);

                    printf("%lld\n",ans);

                }

            else

                {

                    long long data;

                    scanf("%lld",&data);

                    change(l,r,1,1,n,data);

                }

        }

    return 0;

}

BZOJ-3212 Pku3468 A Simple Problem with Integers 裸线段树区间维护查询的相关教程结束。