@[he1234QWQ](/user/228451)
$t$ 数组开 $4$ 倍
by Silence_water @ 2021-02-02 15:44:34
已AC了，谢谢你 @[OIer陈奕莱](/user/338630)
by he1234QWQ @ 2021-02-02 15:48:59
我重开了数组，用线段树1的方式AC了这道题
```c
#include<bits/stdc++.h>
using namespace std;
long long t[2000000],a[2000000],lan[2000000];
void fl(long long x,long long l,long long r)
{
if(lan[x]==0) return ;
long long val=lan[x],mid=(l+r)/2;
lan[x*2]+=val,t[x*2]+=(mid-l+1)*val;
lan[x*2+1]+=val,t[x*2+1]+=(r-mid)*val;
lan[x]=0;
}
void f(long long x,long long l,long long r)
{
if(l==r)
{
t[x]=a[l];
return ;
}
long long mid=(l+r)/2;
f(x*2,l,mid);
f(x*2+1,mid+1,r);
t[x]=t[x*2]+t[x*2+1];
}
long long sum(long long x,long long l,long long r,long long L,long long R)
{
if(L<=l&&r<=R)
return t[x];
long long ans=0;
long long mid=(l+r)/2;
fl(x,l,r);
if(L<=mid)
ans+=sum(x*2,l,mid,L,R);
if(mid<R)
ans+=sum(x*2+1,mid+1,r,L,R);
t[x]=t[x*2]+t[x*2+1];
return ans;
}
void f3(long long x,long long l,long long r,long long L,long long R,long long v)
{
if(L<=l&&r<=R){
lan[x]+=v;
t[x]+=(r-l+1)*v;
return ;
}
fl(x,l,r);
long long mid=(l+r)/2;
if(L<=mid) f3(x*2,l,mid,L,R,v);
if(mid<R) f3(x*2+1,mid+1,r,L,R,v);
t[x]=t[x*2]+t[x*2+1];
}
long long n,m;
int main()
{
cin>>n>>m;
for(long long i=1;i<=n;i++)
scanf("%lld",&a[i]);
f(1,1,n);
for(long long i=1;i<=m;i++)
{
long long p;
scanf("%lld",&p);
if(p==1)
{
long long u,v;
cin>>u>>v;
f3(1,1,n,u,u,v);
}
else
{
long long u,v;
cin>>u>>v;
cout<<sum(1,1,n,u,v)<<endl;
}
}
return 0;
}
```
by he1234QWQ @ 2021-02-02 15:50:07