HDU 6044 Limited Permutation 读入挂+组合数学

Limited Permutation

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

 

题解：

　　看懂题意

　　每个pi掌管 L ,R,题意是指超过这段范围就有比pi还要小的值

　　所有必然有一个pi 值掌管 1，n的，推出  必有  pj，pk分别 掌管 （1，i - 1）， （i+1,n） 

　　dfs下去计算方案

　　还有就是必须用读入挂才能过

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 1e6+10, M = 1e3+20,inf = 2e9,mod = 1e9 + 7;
namespace IO {
    const int MX = 4e7; //1e7占用内存11000kb
    char buf[MX]; int c, sz;
    void begin() {
        c = 0;
        sz = fread(buf, 1, MX, stdin);
    }
    inline bool read(int &t) {
        while(c < sz && buf[c] != '-' && (buf[c] < '0' || buf[c] > '9')) c++;
        if(c >= sz) return false;
        bool flag = 0; if(buf[c] == '-') flag = 1, c++;
        for(t = 0; c < sz && '0' <= buf[c] && buf[c] <= '9'; c++) t = t * 10 + buf[c] - '0';
        if(flag) t = -t;
        return true;
    }
}

const int MOD = (int)1e9 + 7;
int F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘
void init(){
    inv[1] = 1;
    for(int i = 2; i < N; i ++){
        inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;
    }
    F[0] = Finv[0] = 1;
    for(int i = 1; i < N; i ++){
        F[i] = F[i-1] * 1ll * i % MOD;
        Finv[i] = Finv[i-1] * 1ll * inv[i] % MOD;
    }
}
inline LL C(int n, int m){//comb(n, m)就是C(n, m)
    if(m < 0 || m > n) return 0;
    return F[n] * 1ll * Finv[n - m] % MOD * Finv[m] % MOD;
}

struct ss{
    int l,r,id;
    bool operator<(const ss& x) const{
        if(l == x.l) return r > x.r;
        else return l < x.l;
    }
}a[N];
int now,ok;
inline LL dfs(int ll,int rr) {
    if(!ok) return 0;
    if(ll > rr) return 1LL;
    if(a[now].l != ll || a[now].r != rr) {
        ok = 0;
        return 0;
    }
    int ids = a[now++].id;
    return dfs(ll,ids-1) * dfs(ids+1,rr) % mod * C(rr-ll,ids-ll) % mod;
}
int main() {
    init();
    int cas = 1,n;
    IO::begin();
    while(IO::read(n)) {
        for(int i = 1; i <= n; ++i) IO::read(a[i].l);
        for(int i = 1; i <= n; ++i) IO::read(a[i].r),a[i].id = i;
        now = 1, ok = 1;
        sort(a+1,a+n+1);
        printf("Case #%d: %lld
",cas++,dfs(1,n));
    }
    return 0;
}