题意：一个长度为n的序列，T个询问,2种询问1 l r x问[l, r]区间内替换一个数能不能使得gcd[l, r] == x,输出yes和no,2 l x序列第l个数替换为x

题解：区间询问，单点需修改，线段树，算出区间有t个数不是x的倍数,t>1输出no

#include 
     
      #define maxn 501000#define INF 0x3f3f3f3ftypedef long long ll;using namespace std;int gcd[maxn*4];void pushup(int x){    gcd[x] = __gcd(gcd[x<<1], gcd[x<<1|1]);}void build(int l,int r,int x){    if(l == r){        scanf("%d", &gcd[x]);        return ;    }    int mid = (l+r)>>1;    build(l, mid, x<<1);    build(mid+1, r, x<<1|1);    pushup(x);}void update(int l,int r,int x,int t, int c){    if(l == r){        gcd[x] = c;        return ;    }    int mid = (l+r)>>1;    if(t<=mid) update(l, mid, x<<1, t, c);    else update(mid+1, r, x<<1|1, t, c);    pushup(x);}int query(int l,int r,int x,int L,int R,int g){    if(l == r){        return gcd[x]%g?1:0;    }    if(L>=l && R<=r&&gcd[x]%g == 0) return 0;    int mid = l+r>>1;    if(mid < L) return query(mid+1, r, x<<1|1, L, R, g);    else if(mid >= R) return query(l, mid, x<<1, L, R, g);    else{        int b1 = query(l, mid, x<<1, L, mid, g);        if(b1 >= 2) return 2;        int b2 = query(mid+1, r, x<<1|1, mid+1, R, g);        return b1+b2;    }}int main(){    int n, t, a, l ,r, x;    cin>>n;    build(1, n, 1);    scanf("%d", &t);    while(t--){        scanf("%d", &a);        if(a == 1){            scanf("%d%d%d", &l, &r, &x);            if(query(1, n, 1, l, r, x) <= 1) printf("YES\n");            else printf("NO\n");        }        else {            scanf("%d%d", &l, &x);            update(1, n, 1, l, x);        }    }    return 0;}