POJ - 3468A Simple Problem with Integers (线段树区间更新,区间查询和)
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
Hint
#include<iostream>
#include<stdio.h>
using namespace std; const int MAXN = ;
int a[MAXN];
long long int tree[MAXN*];
long long int lazy[MAXN*];
int N,Q;
string s;
int x , y, z;
void push_down(int l ,int r ,int rt)
{
int m = (l+r)/; if(lazy[rt])
{
tree[rt*] += lazy[rt]*(m-l+);
tree[rt*+] += lazy[rt]*(r-m);
lazy[rt*] += lazy[rt];
lazy[rt*+] += lazy[rt];
lazy[rt] = ;
}
}
void bulid_tree(int l ,int r ,int rt)
{
if(l==r)
{
tree[rt] = a[l];
return ;
}
int m = (l+r)/; bulid_tree(l,m,rt*);
bulid_tree(m+,r,rt*+);
tree[rt] = tree[rt*]+tree[rt*+];
} long long int Query(int x ,int y ,int l ,int r ,int rt)
{
long long sum = ;
if(x<=l&&r<=y)
{
return tree[rt];
}
int m = (l+r)/;
push_down(l,r,rt);
if(x<=m)
{
sum += Query(x,y,l,m,rt*);
}
if(m<y)
{
sum += Query(x,y,m+,r,rt*+);
}
return sum;
}
void Update(int x ,int y ,int k ,int l ,int r ,int rt)
{
if(x<=l&&y>=r)
{
tree[rt] += k*(r-l+);
lazy[rt] += k;
return ;
}
push_down(l,r,rt);
int m = (l+r)/;
if(x<=m)
{
Update(x,y,k,l,m,rt*);
}
if(y>m)
{
Update(x,y,k,m+,r,rt*+);
}
tree[rt] = tree[rt*]+tree[rt*+];
}
int main()
{
scanf("%d%d",&N,&Q);
for(int i = ; i <= N;i++)
{
scanf("%d",&a[i]);
}
bulid_tree(,N,);
while(Q--)
{
cin>>s;
if(s[]=='Q')
{
scanf("%d%d",&x,&y);
printf("%lld\n",Query(x,y,,N,)); }
else
if(s[]=='C')
{
scanf("%d%d%d",&x,&y,&z);
Update(x,y,z,,N,);
}
}
return ;
}
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