HDU_1175_莫队+逆元

http://acm.hdu.edu.cn/showproblem.php?pid=5145

初探莫队，就是离线排序后的暴力，但是效率大大提高，中间要除法取模，所以用到了逆元。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define LL long long
#define MOD 1000000007
using namespace std;

int a[],num[],n,m,sizee;
LL ans[],inv[];
struct node
{
    int l,r,num,belong;
}query[];

void init()
{
    for(int i = ;i <= ;i++)
    {
        LL temp = ,x = i;
        int pow = MOD-;
        while(pow)
        {
            if(pow%)    temp = temp*x%MOD;
            x = x*x%MOD;
            pow /= ;
        }
        inv[i] = temp;
    }
}

int cmp(node x,node y)
{
    if(x.belong == y.belong)    return x.r < y.r;
    return x.l<y.l;
}

int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        sizee = sqrt(n);
        for(int i = ;i <= n;i++)    scanf("%d",&a[i]);
        for(int i = ;i <= m;i++)
        {
            scanf("%d%d",&query[i].l,&query[i].r);
            query[i].num = i;
            query[i].belong = (query[i].l-)/sizee+;
        }
        sort(query+,query++m,cmp);
        memset(num,,sizeof(num));
        int ll = ,rr = ;
        LL now = ;
        num[a[]]++;
        for(int i = ;i <= m;i++)
        {
            while(rr < query[i].r)
            {
                rr++;
                num[a[rr]]++;
                now = now*(rr-ll+)%MOD*inv[num[a[rr]]]%MOD;
            }
            while(ll > query[i].l)
            {
                ll--;
                num[a[ll]]++;
                now = now*(rr-ll+)%MOD*inv[num[a[ll]]]%MOD;
            }
            while(ll < query[i].l)
            {
                now = now*num[a[ll]]%MOD*inv[rr-ll+]%MOD;
                num[a[ll]]--;
                ll++;
            }
            while(rr > query[i].r)
            {
                now = now*num[a[rr]]%MOD*inv[rr-ll+]%MOD;
                num[a[rr]]--;
                rr--;
            }
            ans[query[i].num] = now;
        }
        for(int i = ;i <= m;i++)    printf("%lld\n",ans[i]);
    }
    return ;
}