POJ 3468A Simple Problem with Integers(线段树区间更新)

2022-10-01 19:12:17

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 112228		Accepted: 34905
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

题意就是如题，对线段树的区间进行更新，可增可减可修改，并查询区间和，此题就是增，具体实现看代码吧.
AC代码：

Source Code

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

using namespace std;

#define maxsize 100005

#define LL long long

LL num[maxsize];

struct node

{

    LL l,r;

    LL maxn,add;

    int mid()

    {

        return (l+r)/;

    }

} tree[maxsize<<];

void Build_Tree(LL root,LL l,LL r)

{

    tree[root].l=l;

    tree[root].r=r;

    tree[root].add=;

    if(l==r)

    {

        tree[root].maxn=num[l];

        return ;

    }

    Build_Tree(root*,l,tree[root].mid());

    Build_Tree(root*+,tree[root].mid()+,r);

    tree[root].maxn=tree[root*].maxn+tree[root*+].maxn;

}

void Update(LL root,LL l,LL r,LL czp)

{

    if(tree[root].l==l&&tree[root].r==r)

    {

        tree[root].add+=czp;

        return ;

    }

    tree[root].maxn+=((r-l+)*czp);

    if(tree[root].mid()>=r) Update(root<<,l,r,czp);

    else if(tree[root].mid()<l)     Update(root<<|,l,r,czp);

    else

    {

        Update(root<<,l,tree[root].mid(),czp);

        Update(root<<|,tree[root].mid()+,r,czp);

    }

}

long long Query(LL root,LL l,LL r)

{

    if(tree[root].l==l&&tree[root].r==r)

    {

        return tree[root].maxn+(r-l+)*tree[root].add;

    }

    tree[root].maxn+=(tree[root].r-tree[root].l+)*tree[root].add;

    Update(root<<,tree[root].l,tree[root].mid(),tree[root].add);

    Update(root<<|,tree[root].mid()+,tree[root].r,tree[root].add);

    tree[root].add=;

    if(tree[root].mid()>=r) return Query(root*,l,r);

    else if(tree[root].mid()<l) return Query(root*+,l,r);

    else

    {

        LL Lans=Query(root*,l,tree[root].mid());

        LL Rans=Query(root*+,tree[root].mid()+,r);

        return Lans+Rans;

    }

}

int main()

{

    /*ios::sync_with_stdio(false);*/

    char str[];

    LL m,n,a,b,c;

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

    {

        for(LL i=; i<=m; i++)    scanf("%I64d",&num[i]);

        Build_Tree(,,m);

        for(LL i=; i<=n; i++)

        {

            scanf("%s%I64d%I64d",str,&a,&b);

            if(str[]=='Q')    printf("%I64d\n",Query(,a,b));

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

            {

                scanf("%I64d",&c);

                Update(,a,b,c);

            }

        }

    }

    return ;

}

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