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.
InputThe 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.
OutputYou 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. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 200005; long long add[maxn*4],sum[maxn*4]; void pushdown(int rt,int m){ if(add[rt]){ add[rt*2] += add[rt]; add[rt*2+1]+= add[rt]; sum[rt*2] += add[rt]*(m-m/2); sum[rt*2+1]+= add[rt]*(m/2); add[rt]=0; } } void pushup(int rt){ sum[rt]=sum[rt*2]+sum[rt*2+1]; } void build(int l,int r,int rt){ add[rt]=0; if(l == r){ scanf("%lld",&sum[rt]); return ; } int m=(l+r)/2; build(l,m,rt*2); build(m+1,r,rt*2+1); pushup(rt); } void update(int L,int R,int m,int l,int r,int rt){ if(L<=l && R>=r){ add[rt] += m; sum[rt] += (r-l+1)*(long long)m; return ; } pushdown(rt,r-l+1); int mid=(l+r)/2; if(L<=mid) update(L,R,m,l,mid,rt*2); if(R>mid) update(L,R,m,mid+1,r,rt*2+1); pushup(rt); } long long 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 mid = (l+r)/2; long long ans=0; if(L<=mid) ans+=query(L,R,l,mid,rt*2); if(R>mid) ans+=query(L,R,mid+1,r,rt*2+1); return ans; } int main(){ int n,t; int q; while((scanf("%d%d",&n,&q))!=EOF){ build(1,n,1); char ch[5]; while(q--){ cin>>ch; if(ch[0]== 'Q'){ int l,r; scanf("%d%d",&l,&r); printf("%lld\n",query(l,r,1,n,1)); } else{ int l,r,x; scanf("%d%d%d",&l,&r,&x); update(l,r,x,1,n,1); } } } return 0; }