A Simple Problem with Integers POJ - 3468（线段树）

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 A Simple Problem with Integers

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.

思路：线段树区间更新，区间求和

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 1000050
#define ll long long
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
ll sum[N<<2],lazy[N<<2];
void pushup(ll rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(ll rt,ll m)
{
    if(lazy[rt])
    {
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        sum[rt<<1]+=lazy[rt]*(m-(m>>1));
        sum[rt<<1|1]+=lazy[rt]*(m>>1);
        lazy[rt]=0;
    }
}
void build(ll l,ll r,ll rt)
{
    lazy[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);
        return;
    }
    ll m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(ll L,ll R,ll c,ll l,ll r,ll rt)
{
    if(L<=l&&r<=R)
    {
        lazy[rt]+=c;
        sum[rt]+=(ll)c*(r-l+1);
        return;
    }
    pushdown(rt,r-l+1);
    ll m=(l+r)>>1;
    if(L<=m) update(L,R,c,lson);
    if(m<R) update(L,R,c,rson);
    pushup(rt);
}
ll query(ll L,ll R,ll l,ll r,ll rt)
{
    if(L<=l&&r<=R)
        return sum[rt];
    pushdown(rt,r-l+1);
    ll m=(l+r)>>1;
    ll ans=0;
    if(L<=m)
        ans+=query(L,R,lson);
    if(R>m)
        ans+=query(L,R,rson);
    return ans;
}
int main()
{
    ll n,m,a,b,c;
    char s;
    scanf("%lld%lld",&n,&m);
    build(1,n,1);
    while(m--)
    {
        scanf(" %c",&s);
        if(s=='Q')
        {
            scanf("%lld%lld",&a,&b);
            printf("%lld\n",query(a,b,1,n,1));
        }
        else
        {
            scanf("%lld%lld%lld",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
    }
    return 0;
}