poj 3468 區間更新 整個區間加一個數和區間求和操作

life4711發表於2014-08-01

http://poj.org/problem?id=3468

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.
線段樹的裸題，只有對於我這種初學者才會想好久==

幾點注意：1. 線段樹裡面sum值和flag值要用long long

                    2. 輸入時要有一個getchar（）取出回車符

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
typedef long long LL;
const int N=100005;
LL a[N];
struct SegementTree
{
    struct Tree
    {
        int l,r;
        LL sum;
        LL flag;
    }tree[N*4];
    void push_up(int root)
    {
        tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;
    }
    void push_down(int root)
    {
        if(tree[root].flag!=0)
        {
            tree[root<<1].flag+=tree[root].flag;
            tree[root<<1|1].flag+=tree[root].flag;//左右子樹做標記
            tree[root<<1].sum+=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].flag;
            tree[root<<1|1].sum+=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].flag;
            tree[root].flag=0;//該節點的子樹已經更新完畢，其取消標記
        }
    }
    void build(int root,int L,int R)
    {
        tree[root].l=L;
        tree[root].r=R;
        tree[root].flag=0;
        if(tree[root].l==tree[root].r)
        {
            tree[root].sum=a[L];
            return;
        }
        int mid=L+(R-L)/2;
        build(root<<1,L,mid);
        build(root<<1|1,mid+1,R);
        push_up(root);
    }
    void update(int root,int L,int R,int k)
    {
        if(L<=tree[root].l&&tree[root].r<=R)
        {
            tree[root].flag+=k;
            tree[root].sum+=(tree[root].r-tree[root].l+1)*k;
            return;
        }
        push_down(root);//本次更新到該區間，要把上次更新的區間值向其左右子樹更新上次的值
        int mid=tree[root].l+(tree[root].r-tree[root].l)/2;
        if(L<=mid)
            update(root<<1,L,R,k);
        if(mid<R)
            update(root<<1|1,L,R,k);
        push_up(root);
    }
    LL query(int root,int L,int R)
    {
        if(L<=tree[root].l&&tree[root].r<=R)
        {
            return tree[root].sum;
        }
        push_down(root);//查詢到該區間要把上次更新的值向其左右子樹進行更新，保證最後統計的數是全部更新後的值
        LL sum=0;
        int mid=tree[root].l+(tree[root].r-tree[root].l)/2;
        if(L<=mid)
            sum+=query(root<<1,L,R);
        if(mid<R)
            sum+=query(root<<1|1,L,R);
        return sum;
    }
}tr;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        tr.build(1,1,n);
        char c;
        int x,y,k;
        for(int i=0;i<m;i++)
        {
            scanf("\n%c",&c);
            if(c=='Q')
            {
                scanf("%d%d",&x,&y);
                printf("%lld\n",tr.query(1,x,y));
            }
            else if(c=='C')
            {
                scanf("%d%d%d",&x,&y,&k);
                tr.update(1,x,y,k);
            }
        }
    }
    return 0;
}


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