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.

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.

You need to answer all Q commands in order. One answer in a line.

　　1）令delta[s] = delta[s] + d，表示将A[s]...A[n]同时增加d，但这样A[t+1]...A[n]就多加了d，所以

　　2）再令delta[t+1] = delta[t+1] - d，表示将A[t+1]...A[n]同时减d

　　    然后来看查询操作query(s, t)，求A[s]...A[t]的区间和，转化为求前缀和，设sum[i] = A[1]+...+A[i]，则

                            A[s]+...+A[t] = sum[t] - sum[s-1]，

　　那么前缀和sum[x]又如何求呢？它由两部分组成，一是数组的原始和，二是该区间内的累计增量和, 把数组A的原始值保存在数组org中，

　　并且delta[i]对sum[x]的贡献值为delta[i]*(x+1-i)，那么  

　　　　                         sum[x] = org[1]+...+org[x] + delta[1]*x + delta[2]*(x-1) + delta[3]*(x-2)+...+delta[x]*1

                                         = org[1]+...+org[x] + segma(delta[i]*(x+1-i))

                                         = segma(org[i]) + (x+1)*segma(delta[i]) - segma(delta[i]*i)，1 <= i <= x

　　　　　　　　　　=segma(org[i]-delta[i]*i)+(x+1)*delta[i],  i<=1<=x  

　　这其实就是三个数组org[i], delta[i]和delta[i]*i的前缀和，org[i]的前缀和保持不变，事先就可以求出来，delta[i]和delta[i]*i的前缀和是不断变化的，可以用两个树状数组来维护。

 1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector>10 #include <queue>11 #include <map>12 #include <set>13 using namespace std;14 typedef long long LL;15 const int MAXN = 100011;16 int n,m;17 int a[MAXN];18 LL sum[MAXN],c1[MAXN],c2[MAXN],ans;19 char ch[12];20 21 inline int getint()22 {23        int w=0,q=0; char c=getchar();24        while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); if(c==‘-‘) q=1,c=getchar(); 25        while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w;26 }27 28 inline void add(LL x,LL val){29     LL cun=x;30     while(x<=n) {31     c1[x]+=val; c2[x]+=val*cun;32     x+=x&(-x);33     }34 }35 36 inline LL query(LL x){37     LL total=0; LL cun=x;38     while(x>0) {39     total+=c1[x]*(cun+1)-c2[x];40     x-=x&(-x);41     }42     return total;43 }44 45 inline LL count(LL l,LL r){46     return query(r)-query(l-1);47 }48 49 inline void work(){50     n=getint(); m=getint(); for(int i=1;i<=n;i++) a[i]=getint(),sum[i]=sum[i-1]+a[i];51     int l,r; LL val;52     while(m--) {53     scanf("%s",ch);    54     if(ch[0]==‘C‘) {55         l=getint(); r=getint(); val=getint();56         add(l,val); add(r+1,-val);57     } else{58         l=getint(); r=getint();59         ans=sum[r]-sum[l-1]; ans+=count(l,r);60         printf("%lld\n",ans);61     }62     }63 }64 65 int main()66 {67   work();68   return 0;69 }