You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
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
树状数组区间更新, 区间查询. 或使用线段树的延迟标签进行区间修改, 这里使用树状数组
- //#include <bits/stdc++.h>
- #include <cstdio>
- #include <algorithm>
- #include <cstring>
- #include <iostream>
- using namespace std;
- typedef long long ll;
- const int maxn=1e5+5;
- ll tree[maxn][2];
- int n;
- int lowbit(int x)
- {
- return x&-x;
- }
- void update(ll x,int i,int f)
- {
- while(i<=n)
- {
- tree[i][f]+=x;
- i+=lowbit(i);
- }
- }
- ll query(int i,int f)
- {
- ll ans=0;
- while(i>0)
- {
- ans+=tree[i][f];
- i-=lowbit(i);
- }
- return ans;
- }
- ll ask(int i)
- {
- return 1ll*(i+1)*query(i,0)-1ll*query(i,1);
- }
- int main()
- {
- int q;
- scanf("%d%d",&n,&q);
- int x=0,y;
- for(int i = 1;i <= n;++i)
- {
- scanf("%d",&y);
- update(1ll*y-x,i,0);
- update(1ll*(y-x)*i,i,1);
- x=y;
- }
- while(q--)
- {
- char op;
- scanf("%c",&op);
- if(op=='C')
- {
- int a,b,x;
- scanf("%d%d%d",&a,&b,&x);
- update(x,a,0);
- update(-x,b+1,0);
- update(1ll*x*a,a,1);
- update(1ll*-x*(b+1),b+1,1);
- }
- else
- {
- int a,b;
- scanf("%d%d",&a,&b);
- ll ans=ask(b)-ask(a-1);
- printf("%lld\n",ans);
- }
- }
- return 0;
- }
来源: http://www.bubuko.com/infodetail-3035526.html