hdu NPY and girls 莫队+逆元

Posted 2020-09-10 xjhz

NPY and girls

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

NPY‘s girlfriend blew him out!His honey doesn‘t love him any more!However, he has so many girlfriend candidates.Because there are too many girls and for the convenience of management, NPY numbered the girls from 1 to n.These girls are in different classes(some girls may be in the same class).And the i-th girl is in class ai.NPY wants to visit his girls frequently.Each time he visits some girls numbered consecutively from L to R in some order. He can only visit one girl every time he goes into a classroom,otherwise the girls may fight with each other(-_-!).And he can visit the class in any order.
Here comes the problem,(NPY doesn‘t want to learn how to use excavator),he wonders how many different ways there can be in which he can visit his girls.The different ways are different means he visits these classrooms in different order.

 

 

Input

The first line contains the number of test cases T(1≤T≤10).
For each test case,there are two integers n,m(0<n,m≤30000) in the first line.N is the number of girls,and M is the number of times that NPY want to visit his girls.
The following single line contains N integers, a1,a2,a3,…,an, which indicates the class number of each girl. (0<ai≤30000)
The following m lines,each line contains two integers l,r(1≤l≤r≤n),which indicates the interval NPY wants to visit.

 

 

Output

For each visit,print how many ways can NPY visit his girls.Because the ans may be too large,print the ans mod 1000000007.

 

 

Sample Input

2 4 2 1 2 1 3 1 3 1 4 1 1 1 1 1

 

 

Sample Output

3 12 1

 

 

Source

BestCoder Round #22

 

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

思路：莫队；

　　　主要是更新；

　　　开始不知道怎么去重；

　　　然后看的别人的；

　　　c（n+1,m+1)=c(n,m)*(n+1)/(m+1);

　　　不太懂；

　　　不去重，有n+1个位置可以插入；

　　　然后将那么数字全取出来，可以插入的位置为个数+1

　　　相当于*(n+1)/(个数+1）；

　　　trick：逆元需要打表；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define mk make_pair
#define eps 1e-7
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e4+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1000000007;

///   数组大小

int pos[N];
struct is
{
    int l,r,now;
    bool operator < (const is &a)const
    {
        if(pos[l]!=pos[a.l])
            return pos[l]<pos[a.l];
        return r<a.r;
    }
} s[N];

int a[N];
ll flag[N],ans,out[N],si;

ll quick_pow(ll x,ll p)
{
    if(!p) return 1;
    ll ans = quick_pow(x,p>>1);
    ans = ans*ans%mod;
    if(p & 1) ans = ans*x%mod;
    return ans;
}

ll inv[N];
void add(int x)
{
    si++;
    flag[a[x]]++;
    if(si<=0)return;
    ans=(ans*si)%mod;
    ans=(ans*inv[flag[a[x]]])%mod;
}

void del(int x)
{
    if(si>0)ans=(ans*inv[si])%mod,ans=(ans*flag[a[x]])%mod;
    flag[a[x]]--;
    si--;
}

void xjhz()
{
    inv[1] = 1;
    for (int i=2; i<=30000; ++i) {
        inv[i] =  (mod - mod / i) * inv[mod%i] % mod;
    }
}

void init()
{
    ans=1;
    si=0;
    memset(flag,0,sizeof(flag));
}
int main()
{
    xjhz();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n,q;
        scanf("%d%d",&n,&q);
        int k=sqrt(n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]),pos[i]=(i-1)/k+1;
        for(int i=1; i<=q; i++)
            scanf("%d%d",&s[i].l,&s[i].r),s[i].now=i;
        sort(s+1,s+1+q);
        int L=1,R=0;
        for(int i=1; i<=q; i++)
        {
            while(L<s[i].l)
            {
                del(L);
                L++;
            }
            while(L>s[i].l)
            {
                L--;
                add(L);
            }
            while(R>s[i].r)
            {
                del(R);
                R--;
            }
            while(R<s[i].r)
            {
                R++;
                add(R);
            }
            out[s[i].now]=ans;
        }
        for(int i=1; i<=q; i++)
            printf("%lld\n",out[i]);
    }
    return 0;
}