A Simple Problem with Integers

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.

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, ... , 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.

Output

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

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

区间更新，用sum记录每个节点的增量，对于某个区间，什么时候用什么时候更新

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<vector>using namespace std;const int MAXN=100000+10;const int MAXM=1000000+10;const int inf=0x3f3f3f;int x[MAXN];struct node{    int left,right;    long long value;    long long sum;} z[3*MAXN];void build(int i,int left,int right){    z[i].left=left;    z[i].right=right;    z[i].sum=0;    if(left==right)    {        z[i].value=x[left];        return ;    }    int mid=(left+right)/2;    build(i<<1,left,mid);    build((i<<1)+1,mid+1,right);    z[i].value=z[i<<1].value+z[(i<<1)+1].value;}void update(int i,int left,int right,long long num){    if(z[i].left==left&&z[i].right==right)    {        z[i].sum+=num;        return ;    }    z[i].value+=(right-left+1)*num;    int mid=(z[i].left+z[i].right)/2;    if(right<=mid)        update(i<<1,left,right,num);    else if(left>mid)        update((i<<1)+1,left,right,num);    else    {        update(i<<1,left,mid,num);        update((i<<1)+1,mid+1,right,num);    }}long long inquiry(int i,int left,int right){    if(z[i].left==left&&z[i].right==right)        return z[i].value+(right-left+1)*z[i].sum;    z[i].value+=(z[i].right-z[i].left+1)*z[i].sum;    int mid=(z[i].left+z[i].right)/2;    update(i<<1,z[i].left,mid,z[i].sum);    update((i<<1)+1,mid+1,z[i].right,z[i].sum);    z[i].sum=0;    if(right<=mid)        return inquiry(i<<1,left,right);    else if(left>mid)        return inquiry((i<<1)+1,left,right);    else        return inquiry(i<<1,left,mid)+inquiry((i<<1)+1,mid+1,right);}int main(){    int n,m;    string s;    int a,b,c;    while(~scanf("%d%d",&n,&m))    {        for(int i=1; i<=n; i++)            scanf("%d",&x[i]);        build(1,1,n);        while(m--)        {            cin>>s;            if(s=="Q")            {                scanf("%d %d",&a,&b);                printf("%lld\n",inquiry(1,a,b));            }            else if(s=="C")            {                scanf("%d%d%d",&a,&b,&c);                update(1,a,b,c);            }        }    }    return 0;}