題目鏈接:https://www.lydsy.com/JudgeOnline/problem.php?id=5028
思路:因爲gcd(a, b, c) =gcd(a, b − a, c − b),所以可以用線段樹維護差分數組,區間加操作只會改變第一個數的差分值,所以變爲了單點修改操作
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n,m,x,y,op,a[100008],z;
struct as{
int l,r,su,gcd;
}tr[100008<<2];
void push(int d){tr[d].su=tr[lc].su+tr[rc].su;tr[d].gcd=__gcd(tr[lc].gcd,tr[rc].gcd);}
void build(int d,int l,int r)
{
tr[d].l=l,tr[d].r=r;
if(l==r) {tr[d].su=tr[d].gcd=a[l];return;}
int mid=(l+r)>>1;
build(lc,l,mid);
build(rc,mid+1,r);
push(d);
}
void add(int d,int l,int pos)
{
if(tr[d].l==tr[d].r&&tr[d].l==l) {tr[d].gcd+=pos;tr[d].su+=pos;return;}
int mid=(tr[d].l+tr[d].r)>>1;
if(mid>=l) add(lc,l,pos);
else add(rc,l,pos);
push(d);
}
int query(int d,int l,int r,int z)
{
if(l>r) return 0;
if(tr[d].l==l&&tr[d].r==r)
{
if(z) return tr[d].gcd;
else return tr[d].su;
}
int mid=(tr[d].l+tr[d].r)>>1;
if(mid>=r) return query(lc,l,r,z);
else if(l>mid) return query(rc,l,r,z);
else
{
if(z) return __gcd(query(lc,l,mid,z),query(rc,mid+1,r,z));
else return query(lc,l,mid,z)+query(rc,mid+1,r,z);
}
}
int main()
{
cin.tie(0);
cout.tie(0);
cin>>n>>m;
FOR(i,1,n)
{
si(x);
if(i==1) a[i]=x;
else a[i]=x-y;
y=x;
}
build(1,1,n);
while(m--)
{
si(op),si(x),si(y);
if(op==1) printf("%d\n",abs(__gcd(query(1,1,x,0),query(1,x+1,y,1))));
else
{
si(z),add(1,x,z);
if(y+1<=n) add(1,y+1,-z);
}
}
return 0;
}
思路: