![POJ – 3468 A Simple Problem with Integers（[线段树] 区间增加+区间和查询）](http://47.106.118.63/wp-content/uploads/2018/08/光之教堂.jpg)
撰写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
题解：
用add标志标记已经修改但子节点未被修改的节点，在查询时更新需要向下更新的子节点即可
代码：
//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
char c;
int sgn;
if(c=getchar(),c==EOF) return 0; //EOF
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
inline void outt(int x) {
if(x>9) outt(x/10);
putchar(x%10+'0');
}
const int maxn = 1e6+10;
struct SegmentTree{
int l,r;
long long sum, add;
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
}tree[1000010*4];
int a[1000010],n,m;
void build(int p,int l,int r){
l(p)=l,r(p)=r;
if(l==r){
sum(p)=a[l];return ;
}
int mid=(l+r)/2;//只有这里的一半是直接用传入的l，r，但实际上这里的l，r就是下面的p.l,p.r
build(p*2,l,mid);
build(p*2+1,mid+1,r);
sum(p)=sum(p*2)+sum(p*2+1);
}
//延迟标记代表：该节点已经被修改，但子节点尚未被更新
void spread(int p){
if(add(p)){
sum(p*2) += add(p)*(r(p*2)-l(p*2)+1); //更新左子节点信息
sum(p*2+1) += add(p)*(r(p*2+1)-l(p*2+1)+1);//更新右子节点信息
add(p*2) += add(p); //给左子节点打延迟标记
add(p*2+1) += add(p);
add(p) = 0;//清楚p节点的标记
}
}
void change(int p,int l,int r,int d){
if(l<=l(p) && r>=r(p)){//完成覆盖
sum(p)+=(long long)d*(r(p)-l(p)+1);//更新节点信息
add(p)+=d;//给节点打延迟标记
return ;
}
spread(p); //下传延迟标记
int mid=(l(p)+r(p))/2; //这里的中间是当前节点的中间（或者说是偏左的中间）位置
if(l<=mid)change(p*2,l,r,d);
if(r>mid)change(p*2+1,l,r,d);//注意这里不是非此即彼的关系
sum(p)=sum(p*2)+sum(p*2+1);
}
long long ask(int p,int l,int r){
if(l<=l(p) && r>=r(p))return sum(p);
spread(p);
int mid=(l(p)+r(p))/2;
long long val=0;
if(l <= mid) val+=ask(p*2,l,r);//注意！这里对中间元素进行比较的是l不是tree[p].l
if(r > mid) val+=ask(p*2+1,l,r);
return val;
}
char cmd[10];
int main(){
int n,m;
scan_d(n),scan_d(m);
rep(i,1,n)scan_d(a[i]);
build(1,1,n);
while(m--){
scanf("%s", cmd);
int l,r,d;
scan_d(l);
scan_d(r);
if(cmd[0]=='Q'){
printf("%I64d\n", ask(1,l,r));
}else{
scan_d(d);
change(1,l,r,d);
}
}
return 0;
}