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
The sums may exceed the range of 32-bit integers
这个题有的数据要开 long long ,否则会 wa掉,数据较大的话用 scanf、printf 会快一点,
或者用cin、cout但是要加上 std::ios::sync_with_stdio(false); 这样会快一点;
当然,我也不是太理解线段树,有的地方也没有看懂(哎)........
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
#define ls o<<1
#define rs o<<1|1
typedef long long ll;
const int M=1e5+10;
ll a[M];//必须用long long
struct node{
int l,r;
ll x;//必须用long long,否则wrong answer
ll lazy;
}t[M<<2];
inline void push_up(int o){//向上传递
t[o].x =t[ls].x +t[rs].x ;
}
inline void build(int l,int r,int o){
t[o].l =l; t[o].r =r; t[o].lazy =0;
if(l==r) {
t[o].x =a[l];
return ;
}
int mid=(l+r)>>1;
build(l,mid,ls); build(mid+1,r,rs);
push_up(o);//叶节点的上一个
}
inline void push_down(int o,int m){//向下传递
if(t[o].lazy ){
t[ls].lazy +=t[o].lazy ;
t[rs].lazy +=t[o].lazy ;
t[ls].x +=t[o].lazy *(m-m/2);//左边一半
t[rs].x +=t[o].lazy *(m/2);//右边一半
t[o].lazy =0;//标记为0
}
}
inline ll query(int l,int r,int o){
if(l<=t[o].l &&t[o].r <=r){
return t[o].x ;
}
push_down(o,t[o].r -t[o].l +1);
int mid=(t[o].l +t[o].r )>>1;
if(r<=mid) return query(l,r,ls);
else if(l>mid) return query(l,r,rs);
else return query(l,mid,ls)+query(mid+1,r,rs);
}
inline void update(int l,int r,int o,int x){
if(l<=t[o].l&&t[o].r<=r){
t[o].lazy +=x;
t[o].x +=x*(t[o].r -t[o].l +1);
return ;
}
push_down(o,t[o].r -t[o].l +1);//我也不知道为什么加这句
int mid=(t[o].l +t[o].r )>>1;
if(r<=mid) update(l,r,ls,x);
else if(l>mid) update(l,r,rs,x);
else{
update(l,mid,ls,x);
update(mid+1,r,rs,x);
}
push_up(o);
}
int main()
{
int n,q;
// std::ios::sync_with_stdio(false);
while(cin>>n>>q){
memset(a,0,sizeof(a));
memset(t,0,sizeof(t));
for(int i=1;i<=n;i++){
//cin>>a[i];
scanf("%lld",&a[i]);
}
build(1,n,1);
char s[2];
int l,r,x;
while(q--){
scanf("%s",s);
if(s[0]=='Q'){
//cin>>l>>r;
scanf("%d%d",&l,&r);
//cout<<query(l,r,1)<<endl;
printf("%lld\n",query(l,r,1));
}
if(s[0]=='C'){
//cin>>l>>r>>x;
scanf("%d%d%d",&l,&r,&x);
update(l,r,1,x);
}
}
}
return 0;
}