BZOJ3212: Pku3468 A Simple Problem with Integers(线段树)
3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 2530 Solved: 1096
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Description
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
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
55
9
15
HINT
The sums may exceed the range of 32-bit integers.
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=;
int Lazy[maxn]; ll sum[maxn];
void pushup(int Now){ sum[Now]=sum[Now<<]+sum[Now<<|];}
void pushdown(int Now,int L,int R){
if(Lazy[Now]){
int Mid=(L+R)>>;
sum[Now<<]+=(ll)Lazy[Now]*(Mid-L+);
sum[Now<<|]+=(ll)Lazy[Now]*(R-Mid);
Lazy[Now<<]+=Lazy[Now];
Lazy[Now<<|]+=Lazy[Now];
Lazy[Now]=;
}
}
void build(int Now,int L,int R)
{
if(L==R) {
scanf("%lld",&sum[Now]);
return ;
}
int Mid=(L+R)>>;
build(Now<<,L,Mid); build(Now<<|,Mid+,R);
pushup(Now);
}
ll query(int Now,int L,int R,int l,int r)
{
if(l<=L&&r>=R) return sum[Now];
int Mid=(L+R)>>; ll res=;
pushdown(Now,L,R);
if(l<=Mid) res+=query(Now<<,L,Mid,l,r);
if(r>Mid) res+=query(Now<<|,Mid+,R,l,r);
pushup(Now);
return res;
}
void update(int Now,int L,int R,int l,int r,int c)
{
if(l<=L&&r>=R){ sum[Now]+=(ll)(R-L+)*c; Lazy[Now]+=c; return ;}
int Mid=(L+R)>>; pushdown(Now,L,R);
if(l<=Mid) update(Now<<,L,Mid,l,r,c);
if(r>Mid) update(Now<<|,Mid+,R,l,r,c);
pushup(Now);
}
int main()
{
int N,M,L,R,x; char opt[];
scanf("%d%d",&N,&M);
build(,,N);
rep(i,,M){
scanf("%s%d%d",opt,&L,&R);
if(opt[]=='Q') printf("%lld\n",query(,,N,L,R));
else scanf("%d",&x),update(,,N,L,R,x);
}
return ;
}
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