分块搞搞
分成n√块,把询问左界不在同一块中的按左界排序,在的按右界排序
然后暴力查询就好了
还是很简单的
不过因为太久没写了今个儿看到一种很妙的写法把我那一堆if去掉了
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define g getchar()
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
inline ll read(){
ll x=0,f=1;char ch=g;
for(;ch<'0'||ch>'9';ch=g)if(ch=='-')f=-1;
for(;ch>='0'&&ch<='9';ch=g)x=x*10+ch-'0';
return x*f;
}
inline void out(ll x){
int a[25],wei=0;
if(x<0)putchar('-'),x=-x;
for(;x;x/=10)a[++wei]=x%10;
if(wei==0){puts("0");return;}
for(int j=wei;j>=1;--j)putchar('0'+a[j]);
putchar('\n');
}
struct re{int l,r,c;}qes[200005];
int Q,n,L,a[50005],sum[1000005],ans,an[200005];
inline bool cmp(re x,re y){return x.l/L==y.l/L?x.r<y.r:x.l/L<y.l/L;}
int main(){
// freopen("","r",stdin);
// freopen("","w",stdout);
n=read();L=(int)sqrt(n+0.5);
for(int i=1;i<=n;++i)a[i]=read();
Q=read();
for(int i=1;i<=Q;++i)qes[i].l=read(),qes[i].r=read(),qes[i].c=i;
sort(qes+1,qes+1+Q,cmp);
sum[a[1]]=1;ans=1;
for(int i=1,r=1,l=1;i<=Q;++i){
for(;r<qes[i].r;)++r,ans+=(sum[a[r]]++==0);
for(;r>qes[i].r;)ans-=(sum[a[r]]--==1),--r;
for(;l>qes[i].l;)--l,ans+=(sum[a[l]]++==0);
for(;l<qes[i].l;)ans-=(sum[a[l]]--==1),++l;
an[qes[i].c]=ans;
}
for(int i=1;i<=Q;++i)out(an[i]);
return 0;
}