POJ3648 A Simple Problem with Integers(线段树之成段更新。入门题)

A Simple Problem with Integers

Time Limit: 5000MS
Memory Limit: 131072K

Total Submissions: 53169
Accepted: 15897

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.

初学线段树：点这

代码：

#include <cstdio>
using namespace std;
typedef long long ll;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=111111;
ll add[maxn<<2];
ll sum[maxn<<2];

void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void PushDown(int rt,int m)
{
    if(add[rt])
    {
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        sum[rt<<1]+=(m-(m>>1))*add[rt];
        sum[rt<<1|1]+=(m>>1)*add[rt];
        add[rt]=0;
    }
}

void build(int l,int r,int rt)
{
    add[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);//
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        sum[rt]+=(r-l+1)*c;
        add[rt]+=c;
        return ;
    }
    PushDown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
         update(L,R,c,lson);
    if(R>m)
         update(L,R,c,rson);
    PushUp(rt);
}

ll query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
        return sum[rt];
    PushDown(rt,r-l+1);
    int m=(l+r)>>1;
    ll res=0;
    if(L<=m)
        res+=query(L,R,lson);
    if(R>m)
        res+=query(L,R,rson);
    return res;
}

int main()
{
    int N,Q;
    scanf("%d%d",&N,&Q);
    build(1,N,1);
    while(Q--)
    {
        char s[6];
        int a,b;
        scanf("%s%d%d",s,&a,&b);
        if(s[0]=='Q')
            printf("%lld\n",query(a,b,1,N,1));//
        else
        {
            int c;
            scanf("%d",&c);
            update(a,b,c,1,N,1);
        }
    }
    return 0;
}