A Simple Problem with Integers(线段树入门题)

Description

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 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 <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <list>
 #include <iomanip>
 #include <cstdlib>
 #include <sstream>
 using namespace std;
 typedef long long LL;
 const int INF=0x5fffffff;
 const double EXP=1e-;
 const int MS=;
 
 struct node
 {
       int l,r;
       LL sum,inc;   //特别注意这里
       int mid()
       {
             return (l+r)/;
       }
 }nodes[*MS];
 
 void creat(int root,int l,int r)
 {
       nodes[root].l=l;
       nodes[root].r=r;
       nodes[root].sum=;
       nodes[root].inc=;
       if(l==r)
             return ;
       creat(root<<,l,(l+r)/);
       creat(root<<|,(l+r)/+,r);
 }
 
 void insert(int root,int pos,int value)
 {
       if(nodes[root].l==nodes[root].r)
       {
             nodes[root].sum+=value;
             return ;
       }
       nodes[root].sum+=value;
       if(pos<=nodes[root].mid())
             insert(root<<,pos,value);
       else
             insert(root<<|,pos,value);
 }
 
 void add(int root,int a,int b,int c)
 {
       if(nodes[root].l==a&&nodes[root].r==b)
       {
             nodes[root].inc+=c;
             return ;
       }
       nodes[root].sum+=c*(b-a+);
       if(b<=nodes[root].mid())
             add(root<<,a,b,c);
       else if(a>nodes[root].mid())
             add(root<<|,a,b,c);
       else
       {
             add(root<<,a,nodes[root].mid(),c);
             add(root<<|,nodes[root].mid()+,b,c);
       }
 }
 
 LL  query(int root,int a,int b)
 {
       if(nodes[root].l>=a&&nodes[root].r<=b)
             return nodes[root].sum+nodes[root].inc*(nodes[root].r-nodes[root].l+);
       nodes[root].sum+=nodes[root].inc*(nodes[root].r-nodes[root].l+);
       add(root<<,nodes[root].l,nodes[root].mid(),nodes[root].inc);
       add(root<<|,nodes[root].mid()+,nodes[root].r,nodes[root].inc);
       nodes[root].inc=;
       if(b<=nodes[root].mid())
             return query(root<<,a,b);
       else if(a>nodes[root].mid())
             return query(root<<|,a,b);
       else
             return query(root<<,a,nodes[root].mid())+query(root<<|,nodes[root].mid()+,b);
 }
 
 int main()
 {
       int N,Q,x,y,z;
       scanf("%d%d",&N,&Q);
       creat(,,N);
       for(int i=;i<=N;i++)
       {
             scanf("%d",&x);
             insert(,i,x);
       }
       char cmd[MS];
       while(Q--)
       {
             scanf("%s",cmd);
             if(cmd[]=='Q')
             {
                   scanf("%d%d",&x,&y);
                   printf("%lld\n",query(,x,y));
             }
             else
             {
                   scanf("%d%d%d",&x,&y,&z);
                   add(,x,y,z);
             }
       }
       return ;
 }