可以直接裸线段树O(mlogn),也可以直接预处理O(n2)。。鉴于m是n2级别的,还是暴力预处理快一些hh
线段树版
#include<cstdio>
#define ll long long
#define N 1010
int n,m,a[N];
struct node{
int l,r,x;
}tree[N<<2];
inline int read(){
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x;
}
inline int gcd(int x,int y){
return y==0?x:gcd(y,x%y);
}
inline void build(int p,int l,int r){
if(l==r){
tree[p].x=a[l];return;
}
int mid=l+r>>1;
build(p<<1,l,mid);
build(p<<1|1,mid+1,r);
tree[p].x=gcd(tree[p<<1].x,tree[p<<1|1].x);
}
inline int query(int p,int l,int r,int x,int y){
if(x<=l&&r<=y) return tree[p].x;
int mid=l+r>>1,res=0;
if(x<=mid) res=gcd(query(p<<1,l,mid,x,y),res);
if(y>mid) res=gcd(query(p<<1|1,mid+1,r,x,y),res);
return res;
}
int main(){
//freopen("a.in","r",stdin);
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) a[i]=read();
build(1,1,n);
while(m--){
int x=read(),y=read();
printf("%d\n",query(1,1,n,x,y));
}
return 0;
}
预处理版
#include <bits/stdc++.h>
#define N 1010
inline int read(){
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x;
}
int n,m,a[N],dp[N][N];
inline int gcd(int x,int y){
return y==0?x:gcd(y,x%y);
}
int main(){
// freopen("a.in","r",stdin);
n=read();m=read();
for(int i=1;i<=n;++i) a[i]=read();
for(int i=1;i<=n;++i)
for(int j=i;j<=n;++j)
dp[i][j]=gcd(a[j],dp[i][j-1]);
while(m--){
int l=read(),r=read();
printf("%d\n",dp[l][r]);
}
return 0;
}