A Simple Problem with Integers

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.

Source

POJ Monthly--2007.11.25, Yang Yi

线段树的基础区域更新

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<stdlib.h>
  5 #include<algorithm>
  6 #define LL __int64
  7 using namespace std;
  8 const int MAXN=100000+10;
  9 const int INF=0x3f3f3f3f;
 10 int b[MAXN];
 11 struct node
 12 {
 13     int l,r;
 14     LL val,col;
 15     int mid()
 16     {
 17         return (l+r)/2;
 18     }
 19 }a[MAXN*4];
 20
 21 void btree(int step,int l,int r)
 22 {
 23     a[step].l=l;
 24     a[step].r=r;
 25     a[step].col=0;
 26     if(l==r)
 27     {
 28         a[step].val=b[l];
 29         //scanf("%d",&a[step].val); 这样输入会WA
 30         return ;
 31     }
 32     int mid=a[step].mid();
 33     btree(step*2,l,mid);
 34     btree(step*2+1,mid+1,r);
 35     a[step].val=a[step*2].val+a[step*2+1].val;
 36 }
 37
 38 void ptree(int step,int L,int R,LL num)
 39 {
 40     if(L==a[step].l&&a[step].r==R)//找到符合条件的区域就不更新了，记录下要要更新的值
 41     {
 42         a[step].col+=num;
 43         return ;
 44     }
 45
 46     a[step].val+=(R-L+1)*num;//如果不是符合条件的区域，对当前结点更新
 47
 48     int mid=a[step].mid();
 49     if(R<=mid)
 50         ptree(step*2,L,R,num);
 51     else if(L>mid)
 52         ptree(step*2+1,L,R,num);
 53     else
 54     {
 55         ptree(step*2,L,mid,num);
 56         ptree(step*2+1,mid+1,R,num);
 57     }
 58 }
 59
 60 LL qtree(int step,int L,int R)
 61 {
 62     if(L==a[step].l&&a[step].r==R)
 63     {
 64         return a[step].val+a[step].col*(R-L+1);
 65     }
 66
 67     a[step].val+=(a[step].r-a[step].l+1)*a[step].col;//对当前结点更新
 68
 69     int mid=a[step].mid();//对子结点更新
 70     ptree(step*2,a[step].l,mid,a[step].col);
 71     ptree(step*2+1,mid+1,a[step].r,a[step].col);
 72     a[step].col=0;
 73
 74     if(R<=mid)
 75         return qtree(step*2,L,R);
 76     else if(L>mid)
 77         return qtree(step*2+1,L,R);
 78     else
 79         return qtree(step*2,L,mid)+qtree(step*2+1,mid+1,R);
 80 }
 81 int main()
 82 {
 83     //freopen("in.txt","r",stdin);
 84     int n,kase;
 85     char ch;
 86     int star,en,num;
 87     while(scanf("%d %d",&n,&kase)!=EOF)
 88     {
 89         for(int i=1;i<=n;i++)
 90             scanf("%d",&b[i]);
 91         btree(1,1,n);
 92         while(kase--)
 93         {
 94             cin>>ch;
 95             if(ch==‘C‘)
 96             {
 97                 scanf("%d %d %d",&star,&en,&num);//num输入%I64d会RE
 98                 ptree(1,star,en,num);
 99             }
100             else
101             {
102                 scanf("%d %d",&star,&en);
103                 printf("%I64d\n",qtree(1,star,en));
104             }
105         }
106     }
107     return 0;
108 }