标签：nbsp sub int for dtree NPU val numbers value
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
代码如下
#include
#include
typedef long long ll;
const int N=1e5+7;
int n,q;
int a[N];
struct node{ //建树
int l,r,len;//多了一个区间长度len和懒标记
ll sum,lazy;
}tree[N*4];
void pushup(int now){//向上更新sum
tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
void buildtree(int now, int l,int r){
tree[now].l=l;
tree[now].r=r;
tree[now].lazy=0;// 建树时懒标记置零
tree[now].len=r-l+1;
if(l==r){
tree[now].sum=a[l];
return;
}
int mid=(l+r)>>1;
buildtree(now<<1,l,mid);
buildtree(now<<1|1,mid+1,r);
pushup(now);
}
void pushdown(int now){ //向子孙传递懒标记
if(tree[now].lazy){
tree[now<<1].sum+=tree[now].lazy*tree[now<<1].len;
tree[now<<1].lazy+=tree[now].lazy;
tree[now<<1|1].sum+=tree[now].lazy*tree[now<<1|1].len;
tree[now<<1|1].lazy+=tree[now].lazy;
tree[now].lazy=0;
}
}
void add(int now,int l,int r,int v){
int L=tree[now].l,R=tree[now].r;
if(l<=L&&R<=r){//如果现区间已经完全在所要修改的区间以内，则不需要继续向下修改，进行懒标记就好
tree[now].sum+=tree[now].len*v;
tree[now].lazy+=v;
return;
}
pushdown(now);//懒标记由父辈传给子辈
int mid=(L+R)>>1;
if(mid>=l)add(now<<1,l,r,v);//与左儿子有交集
if(mid<r)add(now<<1|1,l,r,v);//与右儿子有交集
pushup(now);
}
ll query(int now,int l,int r){
int L=tree[now].l,R=tree[now].r;
if(l==L&&r==R)return tree[now].sum;//现区间的边界与需要查询的边界刚刚好相同直接返回
pushdown(now);//真正懒的所在，需要用到的时候懒标记才会继续向下传递，add里面并没有将子孙更新完全
int mid=(L+R)>>1;
if(l>mid)return query(now<<1|1,l,r);
else if(r<=mid)return query(now<<1,l,r);
else return query(now<<1,l,mid)+query(now<<1|1,mid+1,r);
}
int main(){
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
buildtree(1,1,n);
while(q--){
char str[3];
scanf("%s",str);
if(str[0]==‘C‘){
int x,y,v;
scanf("%d%d%d",&x,&y,&v);
add(1,x,y,v);
}
else{
int x,y;
scanf("%d%d",&x,&y);
printf("%lld\n",query(1,x,y));
}
}
return 0;
}A - A Simple Problem with Integers （线段树的区间修改与区间查询）
标签：nbsp sub int for dtree NPU val numbers value
原文地址：https://www.cnblogs.com/chuliyou/p/12533523.html
