POJ3468：线段树（区间更新，区间查询）

POJ3468：区间更新，区间查询

题解：模板题，放一个lazy标志

#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int const N = 100000 + 10;
ll n,q;
struct Node
{
	ll l,r;
	ll sum,lazy;
	void updata(ll val){
		sum += (r-l+1)*val;		
		lazy += val;
	}
}node[N<<2];
void push_up(ll id){
	node[id].sum = node[id<<1].sum + node[id<<1|1].sum;
}
void push_down(ll id){
	ll lazy =node[id].lazy;
	if(lazy){
		node[id<<1].updata(lazy);
		node[id<<1|1].updata(lazy);
		node[id].lazy = 0;
	} 
}
void build(ll id,ll l,ll r){
	node[id].l = l,node[id].r = r;
	node[id].lazy = 0;
	if(l == r) scanf("%lld",&node[id].sum);
	else{
		ll mid = (l + r) >> 1;
		build(id<<1,l,mid);
		build(id<<1|1,mid+1,r);
		push_up(id);
	}
}
ll query(ll id,ll L,ll R){//求[l,r]区间所有结点的和
	ll l = node[id].l,r = node[id].r;
	if(L <= l && r <= R)   //直接得到区间的结果，不用lazy往下更新
		return node[id].sum;
	else{
		push_down(id);
		ll mid = (l + r) >> 1;
		ll res = 0;
		if(L <= mid)	res += query(id<<1,L,R);
		if(mid < R)		res += query(id<<1|1,L,R);
		push_up(id);
		return res;
	}
} 
void updata(ll id,ll L,ll R,ll k){   //将[l,r]区间的每个数加上k
	ll l = node[id].l,	r = node[id].r;
	if(L <= l && r <= R){
		node[id].updata(k);
	}else{ 
		ll mid = (l + r) >> 1;
		push_down(id);
		if(L <= mid)	updata(id<<1,L,R,k);
		if(mid < R)		updata(id<<1|1,L,R,k);
		push_up(id);
	}
}
int main(){
	scanf("%lld%lld",&n,&q);
	build(1,1,n);
	char t;
	ll x,y,z;
	for(ll i=1;i<=q;i++){
		scanf(" %c",&t);
		if(t == 'Q'){
			scanf("%lld%lld",&x,&y);
			printf("%lld\n",query(1,x,y));
		}else{
			scanf("%lld%lld%lld",&x,&y,&z);
			updata(1,x,y,z);
		}
	}
	return 0;
}