POJ 3468 A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 106771		Accepted: 33308
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

 

 

线段树练习题 

区间修改 区间查询

#include <cstdio>
#define Max 2000000

struct Tree
{
    long long l,r;
    long long lazy,dis;
}tr[Max];
long long n,m;
void up(long long k)
{
    tr[k].dis=tr[k<<1].dis+tr[k<<1|1].dis;
}
void build(long long k,long long l,long long r)
{
    tr[k].l=l;tr[k].r=r;
    if(l==r)
    {
        scanf("%lld",&tr[k].dis);
        return;
    }
    long long mid=(l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    up(k);
}
void down(long long k)
{
    if(tr[k].l==tr[k].r) return;
    tr[k<<1].lazy+=tr[k].lazy;
    tr[k<<1|1].lazy+=tr[k].lazy;
    tr[k<<1].dis+=tr[k].lazy*(tr[k<<1].r-tr[k<<1].l+1);
    tr[k<<1|1].dis+=tr[k].lazy*(tr[k<<1|1].r-tr[k<<1|1].l+1);
    tr[k].lazy=0;
}
void section_change(long long k,long long l,long long r,long long v)
{
    if(tr[k].l==l&&tr[k].r==r)
    {
        tr[k].dis+=v*(r-l+1);
        tr[k].lazy+=v;
        return;
    }
    if(tr[k].lazy) down(k);
    long long mid=(tr[k].l+tr[k].r)>>1;
    if(l>mid) section_change(k<<1|1,l,r,v);
    else if(r<=mid) section_change(k<<1,l,r,v);
    else section_change(k<<1,l,mid,v),section_change(k<<1|1,mid+1,r,v);
    up(k);
}
long long section_query(long long k,long long l,long long r)
{
    if(tr[k].l==l&&tr[k].r==r) return tr[k].dis;
    if(tr[k].lazy) down(k); 
    long long mid=(tr[k].l+tr[k].r)>>1;
    if(l>mid) return section_query(k<<1|1,l,r);
    else if(r<=mid) return section_query(k<<1,l,r);
    else return section_query(k<<1,l,mid)+section_query(k<<1|1,mid+1,r);
}
int main()
{
    scanf("%lld%lld",&n,&m);
    build(1,1,n);
    char ch[4];
    for(long long x,y,z;m--;)
    {
        scanf("%s",ch+1);
        switch(ch[1])
        {
            case 'C':
            {
                scanf("%lld%lld%lld",&x,&y,&z);
                section_change(1,x,y,z);
                break;
            }
            case 'Q':
            {
                scanf("%lld%lld",&x,&y);
                printf("%lld\n",section_query(1,x,y));
                break;
            }
        }
    }
    return 0;
}