Problem
给出了一个序列,你需要处理如下两种询问。
“C a b c”表示给[a, b]区间中的值全部增加c (-10000 ≤ c ≤ 10000)。
“Q a b” 询问[a, b]区间中所有值的和。
涉及线段树的区间更新
lazy一下,该PushDown则PushDown即可。注意Update与Query时候的处理。
#include <cstdio>
#define L(u) (u<<1)
#define R(u) (u<<1|1)
#define ll long long
using namespace std;
const int mx=100000;
struct Tree {
int lch,rch;
ll sum,add;
}tree[mx<<2];
int d[mx+10],n,m,x,y,c;
char ops;
void Pushup(int u) {
tree[u].sum=tree[L(u)].sum+tree[R(u)].sum;
}
void PushDown(int u) {
if (tree[u].add) {
tree[L(u)].add += tree[u].add;
tree[R(u)].add += tree[u].add;
tree[L(u)].sum += tree[u].add * (tree[L(u)].rch - tree[L(u)].lch + 1);
tree[R(u)].sum += tree[u].add * (tree[R(u)].rch - tree[R(u)].lch + 1);
tree[u].add = 0;
}
}
void Build(int u,int l,int r) {
tree[u].lch=l;
tree[u].rch=r;
tree[u].add=0;
if (l==r) {
tree[u].sum=d[l];
return ;
}
int mid=(l+r)>>1;
Build(L(u),l,mid);
Build(R(u),mid+1,r);
Pushup(u);
}
void Update(int u,int l,int r,ll val) {
if (tree[u].lch==l&&tree[u].rch==r) {
tree[u].sum+=val*(r-l+1);
tree[u].add+=val;
return;
}
if (tree[u].lch==tree[u].rch) return;
PushDown(u);
int mid=(tree[u].lch+tree[u].rch)>>1;
if (l>mid) Update(R(u),l,r,val);
else if (r<=mid) Update(L(u),l,r,val);
else {
Update(L(u),l,mid,val);
Update(R(u),mid+1,r,val);
}
Pushup(u);
}
ll Query(int u,int l,int r) {
if (l==tree[u].lch&&tree[u].rch==r) return tree[u].sum;
PushDown(u);
int mid=(tree[u].lch+tree[u].rch)>>1;
if (l>mid) return Query(R(u),l,r);
else if (r<=mid) return Query(L(u),l,r);
else return Query(L(u),l,mid)+Query(R(u),mid+1,r);
}
int main() {
scanf("%d%d",&n,&m);
for (int i=1;i<=n;++i)
scanf("%d",&d[i]);
Build(1,1,n);
while (m--) {
scanf(" %c",&ops);
if (ops=='Q') {
scanf("%d%d",&x,&y);
printf("%lld\n",Query(1,x,y));
} else {
scanf("%d%d%d",&x,&y,&c);
Update(1,x,y,c);
}
}
return 0;
}