A Simple Problem with Integers POJ - 3468 （区间修改+区间查询）

Posted 2021-02-20 shallow-dream

A Simple Problem with Integers  POJ - 3468 

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.

 

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <string>
  5 #include <vector>
  6 #include <map>
  7 #include <set>
  8 #include <list>
  9 #include <deque>
 10 #include <queue>
 11 #include <stack>
 12 #include <cstdlib>
 13 #include <cstdio>
 14 #include <cmath>
 15 #include <iomanip>
 16 #define ull unsigned long long
 17 #define ll long long
 18 #define pb push_back
 19 #define mem(sum,x) memset(sum,x,sizeof(sum))
 20 #define rep(i,start,end) for(int i=start;i<=end;i++)
 21 #define per(i,end,start) for(int i=end;i>=start;i--)
 22 #define tle ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
 23 using namespace std;
 24 const int mod = 998244353;
 25 const int mxn = 2e5 +7;
 26 int _,n,m,k,t,u,v,w,ans,cnt,ok;
 27 int num[mxn] , last[mxn] , key[mxn] ;
 28 /// string str ;
 29 char str[10] , ch ;
 30 #define lc now<<1
 31 #define rc now<<1|1
 32 struct node {ll val , l , r , sum , lazy ;};node no[mxn<<4];
 33 int lowbit(int x) {return x&-x;};
 34 /// void add(int i,int val){while(i<=n*4) key[i]+=val , i+=lowbit(i) ;}
 35 /// int ask(int x){int ans = 0 ;while(x>0) ans+=key[x] , x-=lowbit(x) ;return ans;}
 36 void pushup(int now){
 37     no[now].val = no[lc].val + no[rc].val ;
 38 }
 39 void updown(int now)
 40 {
 41     /// no[now].val = no[lc].val + no[rc].val;
 42     /// no[now].val = max( no[lc].val , no[rc].val );
 43     if(no[now].lazy)
 44     {
 45         no[lc].lazy += no[now].lazy;
 46         no[rc].lazy += no[now].lazy;
 47         no[lc].val += (no[lc].r-no[lc].l+1)*no[now].lazy;
 48         no[rc].val += (no[rc].r-no[rc].l+1)*no[now].lazy;
 49         no[now].lazy = 0;
 50     }
 51 }
 52 void build(int l,int r,int now)
 53 {
 54     no[now].l = l, no[now].r = r ,no[now].lazy = 0 ,no[now].val = 0 ;
 55     if(l==r){
 56         cin>>no[now].val;return ;
 57     }
 58     int mid = (l+r)>>1;
 59     build(l,mid,lc); build(mid+1,r,rc); pushup(now) ;
 60 }
 61 ll ask(int l,int r,int now)
 62 {
 63     if(l<=no[now].l && r>=no[now].r){
 64         return no[now].val;
 65     }
 66     updown(now);
 67     int mid = (no[now].l+no[now].r)>>1;
 68     ll ans = 0 ;
 69     if(l<=mid) ans+=ask(l,r,lc);
 70     if(r>mid) ans+=ask(l,r,rc);
 71     return ans ;
 72 }
 73 void add(int l,int r,int val,int now)
 74 {
 75     if(l<=no[now].l && no[now].r<=r){
 76         no[now].lazy += val ;
 77         no[now].val += (no[now].r-no[now].l+1)*val;
 78         return ;
 79     }
 80     updown(now);
 81     int mid = (no[now].l+no[now].r)>>1;
 82     if(l<=mid) add(l,r,val,lc);
 83     if(r>mid) add(l,r,val,rc);
 84     pushup(now);
 85 }
 86 int main()
 87 {
 88     tle;
 89     while(cin>>n>>m)
 90     {
 91         build(1,n,1);
 92         while(m--)
 93         {
 94             int u,v;
 95             cin>>ch>>u>>v;
 96             if(ch==‘Q‘) cout<<ask(u,v,1)<<endl;
 97             else {
 98             cin>>k; add(u,v,k,1);
 99             }
100         }
101     }
102 }