思路:吧數組分成sqrt(n)塊 當給出更新的長度大於sqrt(n)時我們去樹狀數組暴力更新每個點,小於時,我們讓lazy[x]+=y表示下標x的倍數的位置每個都+=y,詢問時先算樹狀數組的區間,然後再看lazy是否有過標記,ans+=(10/2-3/2)*lazy【2】意思是【3——10】這個區間2的倍數的個數所給出的貢獻。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=2e5+10;
const int inf=0x7f7f7f7f;
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
int block;
ll a[N],lazy[N];
int n,m;
int lowbit(int x)
{
return x&-x;
}
void update(int x,int k)
{
while(x<=n)
{
a[x]+=k;
x+=lowbit(x);
}
}
ll query(int x)
{
ll ans=0;
while(x>0)
{
ans+=a[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
SIS;
cin>>n>>m;
block=sqrt(n);
while(m--)
{
int op,x,y;
cin>>op>>x>>y;
if(op==1)
{
if(x<=block){
lazy[x]+=y;
}else{
for(int i=x;i<=n;i+=x)
update(i,y);
}
}
else
{
ll ans=query(y)-query(x-1);
for(int i=1;i<=block;i++)
ans+=(y/i-(x-1)/i)*lazy[i];
cout<<ans<<endl;
}
}
return 0;
}