BZOJ 2038: [2009国家集训队]小Z的袜子(hose)&&莫队算法

#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=2e5+5;
typedef long long ll;
struct block{int l,r,id;}Q
;
int n,m,bsize,a
,f
;
ll nowans,gg,ans1
,ans2
;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
bool operator <(const block &a,const block &b){
return a.l/bsize!=b.l/bsize?a.l/bsize<b.l/bsize:a.r<b.r;
}
inline void ins(int x){
nowans+=f[x]++;
}
inline void del(int x){
nowans-=--f[x];
}
int main(){
n=read();m=read();bsize=sqrt(n+0.5);
for(int i=1;i<=n;i++) a[i]=read();
for(int i=1;i<=m;i++) Q[i].l=read(),Q[i].r=read(),Q[i].id=i;
sort(Q+1,Q+m+1);
int l=1,r=0;
for(int i=1;i<=m;i++){
while(l>Q[i].l) ins(a[--l]);
while(l<Q[i].l) del(a[l++]);
while(r<Q[i].r) ins(a[++r]);
while(r>Q[i].r) del(a[r--]);
ans1[Q[i].id]=nowans;
ans2[Q[i].id]=1LL*(r-l+1)*(r-l)/2;
}
for(int i=1;i<=m;i++){
if(ans1[i]) gg=__gcd(ans1[i],ans2[i]),printf("%lld/%lld\n",ans1[i]/gg,ans2[i]/gg);
else puts("0/1");
}
return 0;
}

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int N=2e5+7;
struct node{
int l,r,t,pos;
bool operator < (const node &a)const{
return pos==a.pos?r<a.r:pos<a.pos;
}
}b
;
int n,m,l,r,a
,f
;
ll res,ans1
,ans2
;
ll gcd(ll a,ll b){
if(!b) return a;
return gcd(b,a%b);
}
int main(){
//freopen("hose.in","r",stdin);
//freopen("hose.out","w",stdout);
n=read();m=read();
for(int i=1;i<=n;i++) a[i]=read();
int k=sqrt(n*1.0)+0.5;
for(int i=1;i<=m;i++){
b[i].l=read();b[i].r=read();
b[i].t=i;b[i].pos=b[i].l/k;
}
sort(b+1,b+m+1);
memset(f,0,sizeof f);
l=1;r=0;res=0;
for(int i=1;i<=m;i++){
while(r>b[i].r){
res-=(ll)f[a[r]]-1;
f[a[r]]--;
r--;
}
while(r<b[i].r){
r++;
f[a[r]]++;
res+=(ll)f[a[r]]-1;
}
while(l>b[i].l){
l--;
f[a[l]]++;
res+=(ll)f[a[l]]-1;
}
while(l<b[i].l){
res-=(ll)f[a[l]]-1;
f[a[l]]--;
l++;
}
ans1[b[i].t]=res;
ans2[b[i].t]=(ll)(r-l+1)*(r-l)/2;
}
for(int i=1;i<=m;i++){
if(!ans1[i]){
puts("0/1");
continue;
}
ll gg=gcd(ans1[i],ans2[i]);
printf("%lld/%lld\n",ans1[i]/gg,ans2[i]/gg);
//printf("%I64d/%I64d\n",ans1[i]/gg,ans2[i]/gg);
}
return 0;
}