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

31 1 31 3 351 2 2 4 55 2 5 5 5

 

Sample Output

Case #1: 2Case #2: 3

 

Source

2017 Multi-University Training Contest - Team 1

 

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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);    }}