题意：给你n个数，有对区间的加减操作，问某个区间的和是多少。

思路：状压+线段树（要用lazy标记，否则会TLE）

//By SiriusRen

#include <cstdio>

#include <cstring>

#define N 100001

using namespace std;

long long tree[N*4],lazy[N*4];

int xx,yy,zz,n,q;

char jy;

void push_down(int pos,int num){

    lazy[pos*2]+=lazy[pos],lazy[pos*2+1]+=lazy[pos];

    tree[pos*2]+=(num-num/2)*lazy[pos],tree[pos*2+1]+=(num/2)*lazy[pos];

    lazy[pos]=0;

}

void build(int l,int r,int pos){

    if(l==r){scanf("%lld",&tree[pos]);return;}

    int mid=(l+r)/2;

    build(l,mid,pos*2),build(mid+1,r,pos*2+1);

    tree[pos]=tree[pos*2]+tree[pos*2+1];

}

void add(int l,int r,int pos){

    if(l>=xx&&r<=yy){lazy[pos]+=zz;tree[pos]+=(long long)(r-l+1)*zz;return;}

    if(lazy[pos])push_down(pos,r-l+1);

    int mid=(l+r)/2;

    if(mid>=xx)add(l,mid,pos*2);

    if(mid<yy)add(mid+1,r,pos*2+1);

    tree[pos]=tree[pos*2]+tree[pos*2+1];

}

long long query(int l,int r,int pos){

    if(l>=xx&&r<=yy)return tree[pos];

    if(lazy[pos])push_down(pos,r-l+1);

    int mid=(l+r)/2;

    if(mid<xx)return query(mid+1,r,pos*2+1);

    else if(mid>=yy)return query(l,mid,pos*2);

    else return query(l,mid,pos*2)+query(mid+1,r,pos*2+1);

}

int main(){

    scanf("%d%d",&n,&q);

    build(1,n,1);

    while(q--){

        scanf("\n%c%d%d",&jy,&xx,&yy);

        if(jy=='C')scanf("%d",&zz),add(1,n,1);

        else printf("%lld\n",query(1,n,1));

    }

}