CF914D Bash and a Tough Math Puzzle [线段树]-优快云博客

本文链接：https://blog.youkuaiyun.com/sslz_fsy/article/details/95352789

博客介绍了一种区间查找策略，若区间的gcd不是x的倍数，就暴力向下找，当个数大于1时停止查找并输出NO，此方法最多向下找两次，复杂度合理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

如果区间的gcd不是x的倍数, 就暴力向下找, 如果个数 > 1, 就可以不用找了, 并且输出 NO

这样最多向下找两次, 复杂度是正确的

#include<bits/stdc++.h>
#define N 500050
using namespace std;
int read(){ int x; scanf("%d", &x); return x;}
int val[N<<2], a[N], n, m, cnt;
int gcd(int a, int b){ return !b ? a : gcd(b, a % b);}
void Pushup(int x){ val[x] = gcd(val[x<<1], val[x<<1|1]);}
void Build(int x, int l, int r){
	if(l == r){ val[x] = a[l]; return;} int mid = (l+r) >> 1;
	Build(x<<1, l, mid); Build(x<<1|1, mid+1, r); Pushup(x);
}
void Modify(int x, int l, int r, int pos, int v){
	if(l == r){ val[x] = v; return;} int mid = (l+r) >> 1;
	if(pos <= mid) Modify(x<<1, l, mid, pos, v);
	else Modify(x<<1|1, mid+1, r, pos, v); Pushup(x);
}
void Quary(int x, int l, int r, int L, int R, int v){
	if(cnt > 1) return; if(l == r){ cnt++; return;}
	int mid = (l+r) >> 1;
	if(L<=mid &&  val[x<<1] % v) Quary(x<<1, l, mid, L, R, v);
	if(R>mid && val[x<<1|1] % v) Quary(x<<1|1, mid+1, r, L, R, v);
}
int main(){
	n = read();
	for(int i=1; i<=n; i++) a[i] = read();
	Build(1, 1, n);
	m = read();
	while(m--){
		int op = read();
		if(op == 1){
			int l = read(), r = read(), x = read();
			cnt = 0; Quary(1, 1, n, l, r, x);
			if(cnt <= 1) printf("YES\n");
			else printf("NO\n");
		}
		if(op == 2){
			int p = read(), v = read(); Modify(1, 1, n, p, v);
		}
	} return 0;
}