POJ 3468 A Simple Problem with Integers 线段树区间更新 （值在基础上相加或相减）

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

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题意：


思路：线段树区间更新的模板，lazy策略：延迟标记 (或者说懒惰标记), 简单来说就是每次更新的时候不要更新到底,
用延迟标记使得更新延迟到下次需要更新或者询问的时候。
#include <iostream>
#include <cstdio>
#include <map>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
const int maxn = 111111;
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
ll add[maxn<<2];
ll sum[maxn<<2];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
if(add[rt]){
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=add[rt]*(m-(m>>1));
sum[rt<<1|1]+=add[rt]*(m>>1);
add[rt]=0;
}
}
void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r){
scanf("%lld",&sum[rt]);
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if(L<=l&&r<=R){
add[rt]+=c;
sum[rt]+=(ll)c*(r-l+1);
return ;
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) update(L,R,c,lson);
if(m<R) update(L,R,c,rson);
pushup(rt);
}
ll query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R){
return sum[rt];
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
ll ret=0;
if(L<=m) ret+=query(L,R,lson);
if(m<R) ret+=query(L,R,rson);
return ret;
}
int main()
{
int N,Q;
scanf("%d%d",&N,&Q);
build(1,N,1);
while(Q--){
char op[2];
int a,b,c;
scanf("%s",op);
if(op[0]=='Q'){
scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,1,N,1));
}
else{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,N,1);
}
}
return 0;

}