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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 116441		Accepted: 36178
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 題目大意：給N個數，對應編號1~N。對應兩種操作：①求編號a-b對應數字的和。②編號a-b對應數字都加c。 解題思路：線段樹模板，用lazy思想，關鍵是要弄懂pushdown函數。理解了的話這類題目都可以看成是模板題。。
個人對pushdown函數的理解：先用lazy[num]存放的是這個區間上所有的數應該變換的值，即編號為num的結點上的數實際值應該是當前值加上lazy[num]。於是對應左右結點對應的區間的和應該加上lazy[num]*對應長度。然後將lazy傳給左右結點。當要使用到時再調用pushdown。（這個看個人理解。。這種想法可能只適合我自己。。。做題的時候弄懂的）

#include <cstdio>
#include 
 
<algorithm>
using namespace std;
const int N=100005;
const int maxn = N*3;
long long a[maxn],val[maxn],lazy[maxn];
int n,m,ql,qr,A,B,value;

void pushdown(int num,int l)
{
    if(lazy[num])
    {
        lazy[num*2] += lazy[num];
        lazy[num*2+1] += lazy[num];

        val[num*2] += (long long)lazy[num]*(l-(l/2));
        val[num*2+1] += (long long)lazy[num]*(l/2);

        lazy[num] = 0;
    }
}

void build(int num,int l,int r)
{
    if(l==r)
    {
        val[num] = a[l];
        return ;
    }
    int mid = (l+r)/2;
    build(num*2,l,mid);
    build(num*2+1,mid+1,r);
    val[num] = val[num*2]+val[num*2+1];
}

long long Findsum(int num,int l,int r)
{
    int mid = (l+r)/2;
    long long ans = 0;
    if(ql<=l&&qr>=r)
    {
        return val[num];
    }
    else
    {
        pushdown(num,r-l+1);
        if(mid>=ql)
        {
            ans += Findsum(num*2,l,mid);
        }
        if(mid<qr)
        {
            ans += Findsum(num*2+1,mid+1,r);
        }
    }
    return ans;
}

void update(int num,int l,int r)
{
    if(A<=l&&B>=r)
    {
        lazy[num] += (long long)value;
        val[num] += (long long)value*(r-l+1);
        return ;
    }
    pushdown(num,r-l+1);
    int mid = (l+r)/2;
    if(A<=mid)
        update(num*2,l,mid);
    if(B>mid)
        update(num*2+1,mid+1,r);
    val[num] = val[num*2]+val[num*2+1];
}

int main()
{
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
    }
    build(1,1,n);
    char c;
    for(int i=1;i<=m;i++)
    {
        getchar();
        scanf("%c",&c);
        if(c==‘Q‘)
        {
            scanf("%d %d",&ql,&qr);
            printf("%lld\n",Findsum(1,1,n));
        }
        else
        {
            scanf("%d %d %d",&A,&B,&value);
            update(1,1,n);
        }
    }
}

POJ-3468 A Simple Problem with Integers（線段樹、段變化+段查詢、模板）