# A Simple Problem with Integers POJ – 3468(线段树的区间更新)

 Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 118228 Accepted: 36755 Case Time Limit: 2000MS

Description

You have N integers, A1A2, … , 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 A1A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
“C abc” means adding c to each of AaAa+1, … , Ab. -10000 ≤ c ≤ 10000.
“Q ab” means querying the sum of AaAa+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

The sums may exceed the range of 32-bit integers.

#include<iostream>
#include<cstring>
typedef long long ll;
using namespace std;
ll aa[254299];
ll a[600007];
ll lazy[600007];
void build(int n,int l,int r){
if(l==r){
a[n]=aa[l];
return;
}
build(n<<1,l,(l+r)/2);
build(n<<1|1,(l+r)/2+1,r);
a[n]=a[n<<1]+a[n<<1|1];

}

void update(int n,int l,int r,int L,int R ,ll N){
if(L<=l&&R>=r){
a[n]+=N*(r-l+1);
lazy[n]+=N;
return;
}
if(l==r){
return;
}
if(lazy[n]!=0){
update(n<<1,l,(l+r)/2,l,(l+r)/2,lazy[n]);
update((n<<1)+1,(l+r)/2+1,r,(l+r)/2+1,r,lazy[n]);
lazy[n]=0;
}

if(L<=(l+r)/2){

update(n<<1,l,(l+r)/2,L,R,N);
}
if(R>(l+r)/2){
update(n<<1|1,(l+r)/2+1,r,L,R,N);
}
a[n]=a[n<<1]+a[n<<1|1];
}
ll query(int n,int l,int r,int x,int y){
//	cout<<l<<" "<<r<<endl;
if(x<=l&&y>=r){
return a[n];
}

if(lazy[n]!=0){
update(n<<1,l,(l+r)/2,l,(l+r)/2,lazy[n]);
update(n<<1|1,(l+r)/2+1,r,(l+r)/2+1,r,lazy[n]);
lazy[n]=0;
}

ll sum=0;
if(x<=(l+r)/2){
sum+=query(n<<1,l,(l+r)/2,x,y);
}
if(y>(l+r)/2){
sum+=query(n<<1|1,(l+r)/2+1,r,x,y);
}
return sum;

}

int main(){
ios::sync_with_stdio(0);

int n,q;
while(cin>>n>>q){
int k=0;

while((1<<k)<n){
k++;
}

memset(aa,0,sizeof(aa));
memset(a,0,sizeof(a));
memset(lazy,0,sizeof(lazy));
for(int i=1;i<=n;i++)
cin>>aa[i];
build(1,1,1<<k);

char c;
int x,y,z;
for(int i=0;i<q;i++){
cin>>c;

if(c=='Q'){
cin>>x>>y;
cout<<query(1,1,1<<k,x,y)<<endl;
}else{
cin>>x>>y>>z;
update(1,1,1<<k,x,y,z);
}

}

}

return 0;
}