BZOJ 2038 小Z的袜子(hose) （莫队离线）

题目地址：BZOJ 2038
裸的莫队算法。
代码如下：

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
#include <time.h>
using namespace std;
#define LL long long
#define pi acos(-1.0)
#pragma comment(linker, "/STACK:1024000000")
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const double eqs=1e-9;
const int MAXN=50000+10;
int a[MAXN], ha[MAXN];
LL ans1[MAXN], ans2[MAXN];
struct node
{
        int l, r, id, pos;
}fei[MAXN];
LL gcd(LL x, LL y)
{
        return y==0?x:gcd(y,x%y);
}
bool cmp(node x, node y)
{
        return x.pos<y.pos||(x.pos==y.pos&&x.r<y.r);
}
int main()
{
        int n, m, i, j, l, r, k;
        LL res, Gcd;
        while(scanf("%d%d",&n,&m)!=EOF){
                for(i=1;i<=n;i++){
                        scanf("%d",&a[i]);
                }
                k=sqrt(n*1.0)+0.5;
                for(i=0;i<m;i++){
                        scanf("%d%d",&fei[i].l,&fei[i].r);
                        fei[i].id=i;
                        fei[i].pos=fei[i].l/k;
                }
                sort(fei,fei+m,cmp);
                memset(ha,0,sizeof(ha));
                l=1;
                r=0;
                res=0;
                for(i=0;i<m;i++){
                        while(r>fei[i].r){
                                res-=(LL)ha[a[r]]-1;
                                ha[a[r]]--;
                                r--;
                        }
                        while(r<fei[i].r){
                                r++;
                                ha[a[r]]++;
                                res+=(LL)ha[a[r]]-1;
                        }
                        while(l>fei[i].l){
                                l--;
                                ha[a[l]]++;
                                res+=(LL)ha[a[l]]-1;
                        }
                        while(l<fei[i].l){
                                res-=(LL)ha[a[l]]-1;
                                ha[a[l]]--;
                                l++;
                        }
                        ans1[fei[i].id]=res;
                        ans2[fei[i].id]=(LL)(r-l+1)*(r-l)/2;
                }
                for(i=0;i<m;i++){
                        if(!ans1[i]){
                                puts("0/1");
                                continue;
                        }
                        Gcd=gcd(ans1[i],ans2[i]);
                        printf("%lld/%lld\n",ans1[i]/Gcd,ans2[i]/Gcd);
                }
        }
        return 0;
}