HDU 6044 Limited Permutation （组合数+逆元）

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 10^9+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⋅10^6.

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


思路

PS：因为我们要在区间 $[L,R]$ 中选取 $i-L$ 个数放置在 $i$ 的左边，其中区间 $[L,R]$ 除去 $i$ 后剩 $R-L$ 个数，所以为 $C_{R-L}^{i-L}$ 。

AC 代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<stdlib.h>
#include<iostream>
#include<map>
using namespace std;
typedef pair<int,int> P;
typedef __int64 LL;
map<P,int> mapp;

const int maxn = 1e6+10;
const int mod = 1e9+7;
int l[maxn],r[maxn];
LL ans;
LL mul[maxn];
LL inv[maxn];

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++;
}

template <class T>
inline bool scan_d(T & 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;
}

void init()
{
mul[0]=1;
for(int i=1; i<maxn; i++)
mul[i]=(mul[i-1]*i)%mod;

inv[0]=inv[1]=1;
for(int i=2; i<maxn; i++)
inv[i]=(LL)(mod-mod/i)*inv[mod%i]%mod;
for(int i=1; i<maxn; i++)
inv[i]=(inv[i-1]*inv[i])%mod;
}

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

void dfs(int ll,int rr)
{
if(ans==0||rr<ll)return;
int i=mapp[P(ll,rr)];
if(i==0)
{
ans=0;
return;
}
if(ll==rr)return;
int len=rr-ll;
ans=(ans*C(len,i-ll))%mod;
dfs(ll,i-1);
dfs(i+1,rr);
}

int main()
{
int n;
init();
for(int ti=1; scan_d(n); ti++)
{
mapp.clear();
ans=1;
for(int i=1; i<=n; i++)
scan_d(l[i]);
for(int i=1; i<=n; i++)
{
scan_d(r[i]);
mapp[P(l[i],r[i])]=i;
}
dfs(1,n);
printf("Case #%d: %I64d\n",ti,ans);
}
return 0;
}