线段树——D

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.

ps:怕了怕了，数组开大点。空格%c注意，用long long 吧。呜呜呜

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=500005;
ll a[N];
ll sum[N],lazy[N];
void build(ll l,ll r,ll rt){
    if(l==r){
        sum[rt]=a[l];
        return;
    }
    ll mid=(l+r)/2;
    build(l,mid,rt*2);
    build(mid+1,r,rt*2+1);
    sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void pushdown(ll l,ll r,ll rt){
    if(lazy[rt]){
        lazy[rt*2]+=lazy[rt];
        lazy[rt*2+1]+=lazy[rt];
        sum[rt*2]+=lazy[rt]*l;//这里以后还是这样写吧。。。
        sum[rt*2+1]+=lazy[rt]*r;
        lazy[rt]=0;
    }
}
void update(ll l1,ll r1,ll v,ll l,ll r,ll rt){
    if(l1<=l&&r1>=r){
        sum[rt]+=v*(r-l+1);
        lazy[rt]+=v;
        return;
    }
    ll mid=(l+r)/2;
    pushdown(mid-l+1,r-mid,rt);
    if(l1<=mid){
        update(l1,r1,v,l,mid,rt*2);
    }
    if(r1>mid){
        update(l1,r1,v,mid+1,r,rt*2+1);
    }
    sum[rt]=sum[rt*2]+sum[rt*2+1];
}
ll query(ll l1,ll r1,ll l,ll r,ll rt){
    if(l1<=l&&r1>=r){
        return sum[rt];
    }
    ll ret=0;
    ll mid=(l+r)/2;
    pushdown(mid-l+1,r-mid,rt);
    if(l1<=mid){
        ret+=query(l1,r1,l,mid,rt*2);
    }
    if(r1>mid){
        ret+=query(l1,r1,mid+1,r,rt*2+1);
    }
    return ret;
}
int main()
{
    ll n,q;
    scanf("%lld%lld",&n,&q);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }
    build(1,n,1);
    while(q--){
        char s;
        ll a,b,c;
        scanf(" %c",&s);
        if(s=='Q'){
            scanf("%lld %lld",&a,&b);
            ll ans=query(a,b,1,n,1);
            printf("%lld
",ans);
        }
        else{
            scanf("%lld %lld %lld",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
    }
    return 0;
}