版权声明：https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/88249716
题目链接
有两种操作：
1、询问[l, r]区间内改变一个数，是否可以将它的gcd()的值变成x；
2、将i位置上的值改变成y。
然后怎么去处理gcd()，因为可以改变一个，那么线段树去查询这个区间最少要改变的数的个数不就可以了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 5e5 + 7;
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
int N, M, tree[maxN<<2], sum;
inline void pushup(int rt) { tree[rt] = gcd(tree[lsn], tree[rsn]); }
void buildTree(int rt, int l, int r)
{
if(l == r) { scanf("%d", &tree[rt]); return; }
int mid = HalF;
buildTree(Lson);
buildTree(Rson);
pushup(rt);
}
void update(int rt, int l, int r, int qx, int val)
{
if(l == r) { tree[rt] = val; return; }
int mid = HalF;
if(qx <= mid) update(Lson, qx, val);
else update(Rson, qx, val);
pushup(rt);
}
void query(int rt, int l, int r, int ql, int qr, int val)
{
if(sum > 1) return;
if(l == r) { sum++; return; }
int mid = HalF;
if(ql <= mid && tree[lsn] % val) query(QL, val);
if(qr > mid && tree[rsn] % val) query(QR, val);
}
int main()
{
scanf("%d", &N);
buildTree(1, 1, N);
scanf("%d", &M);
int op, x, y, z;
while(M--)
{
scanf("%d%d%d", &op, &x, &y);
if(op == 1)
{
scanf("%d", &z);
sum = 0;
query(1, 1, N, x, y, z);
printf(sum <= 1 ? "YES\n":"NO\n");
}
else update(1, 1, N, x, y);
}
return 0;
}