HDU6044 Limited Permutation (递归,预处理阶乘逆元)

Limited Permutation

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 680    Accepted Submission(s): 130

Problem Description

As to a permutation  p1,p2,⋯,pn  from  1  to  n , it is uncomplicated for each  1≤i≤n  to calculate  (li,ri)  meeting the condition that  min(pL,pL+1,⋯,pR)=pi  if and only if  li≤L≤i≤R≤ri  for each  1≤L≤R≤n .

Given the positive integers  n ,  (li,ri)   (1≤i≤n) , you are asked to calculate the number of possible permutations  p1,p2,⋯,pn  from  1  to  n , meeting the above condition.

The answer may be very large, so you only need to give the value of answer modulo  109+7 .

 

Input

The input contains multiple test cases.

For each test case:

The first line contains one positive integer  n , satisfying  1≤n≤106 .

The second line contains  n  positive integers  l1,l2,⋯,ln , satisfying  1≤li≤i  for each  1≤i≤n .

The third line contains  n  positive integers  r1,r2,⋯,rn , satisfying  i≤ri≤n  for each  1≤i≤n .

It's guaranteed that the sum of  n  in all test cases is not larger than  3⋅106 .

Warm Tips for C/C++: input data is so large (about 38 MiB) that we recommend to use  fread() for buffering friendly.

size_t fread(void *buffer, size_t size, size_t count, FILE *stream); // reads an array of count elements, each one with a size of size bytes, from the stream and stores them in the block of memory specified by buffer; the total number of elements successfully read is returned.

 

Output

For each test case, output " Case # x :  y " in one line (without quotes), where  x  indicates the case number starting from  1  and  y  denotes the answer of corresponding case.

 

Sample Input

  

   3
1 1 3
1 3 3
5
1 2 2 4 5
5 2 5 5 5
  

 

Sample Output

  

   Case #1: 2
Case #2: 3
  

 

Source

2017 Multi-University Training Contest - Team 1

 

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liuyiding   |   We have carefully selected several similar problems for you:   6044  6043  6042  6041  6040 

 

对于一个1-n的排列,对每个位置给出一个l[i]和一个r[i],该位置的数要满足在这段区间内是最小值.若范围超多这段区间,则不是最小值.

考虑L,R这一段区间,先找到覆盖整个区间的那对l[i]和r[i],它的位置一定是这段区间内的值的最小值.

递归地考虑左区间和右区间,整个区间的解的个数为左区间解的个数*右区间解的个数*乘以C(区间长度,左区间长度)

预处理组合数的时候一定不能用递归的方法!!!!!!!!

T到怀疑人生

#include <cstring>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <map>
#include<time.h>
using namespace std;

const int MAXN=1000010;
const int mod=1e9+7;
typedef long long LL;

inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}

inline bool rea(int & x)
{
    char c=nc();x=0;
    if(c==EOF)    return false;
    for(;c>'9'||c<'0';c=nc());
    for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());
    return true;
}
inline bool rea(LL & x)
{
    char c=nc();x=0;
    if(c==EOF)    return false;
    for(;c>'9'||c<'0';c=nc());
    for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());
    return true;
}



LL inv[MAXN];
LL fac[MAXN];

LL Com(int n,int m){
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}

void init(){
    inv[0]=fac[0]=1;
    inv[1]=1;
    for(int i=1;i<MAXN;i++){
        fac[i]=fac[i-1]*i%mod;
    }
    inv[1]=1;
    for(int i=2;i<MAXN;i++){
        inv[i]=(LL)(mod-mod/i)*inv[mod%i]%mod;
    }
    inv[0]=1;
    for(int i=1;i<MAXN;i++){
        inv[i]=inv[i-1]*inv[i]%mod;
    }
}

typedef pair<int,int> P;
map<P,int> mp;
int l[MAXN],r[MAXN];
LL res=1;
void dfs(int L,int R){
    if(res==0)
        return;
    if(R<L)
        return;
    int x=mp[P(L,R)];
    if(x==0){
        res=0;
        return ;
    }
    if(L==R)
        return;
    int len=R-L;
    int tt=x-L;
    res=res*Com(len,tt)%mod;
    dfs(L,x-1);
    dfs(x+1,R);
}

int n;
bool read()
{
    bool res=rea(n);
    if(res==false){
        return false;
    }
    for(int i=1;i<=n;i++)
        rea(l[i]);
    for(int i=1;i<=n;i++)
        rea(r[i]);
    return true;
}

int main(){
    init();
    int cas=1;
    while(read()){
        int ok=1;
        res=1;
        mp.clear();
        for(int i=1;i<=n;i++){
            mp[P(l[i],r[i])]=i;
        }
        dfs(1,n);
        printf("Case #%d: %lld\n",cas++,res);
    }
}