hdu 6044 Limited Permutation

题意：给你n个限制，让你求长度为n的排列列有多少种。每个限制描述为，从li 到 ri这段区间，ai为最小值。

分析：对于一个描述为[1，n] 的限制，它所对应的位置一定是1到n的最小值也就是1，那么这个位置就把区间分成了两份。假设这个位置左边有k个，然后从剩下的n-1个数中选出k个放在左边，都是符合题意的。然后把两个小区间做同样的处理。就得到了答案。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#define mo 1000000007typedef long long ll;namespace fastIO {    #define BUF_SIZE 100000    //fread -> read    bool IOerror = 0;    inline char nc() {        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;        if(p1 == pend) {            p1 = buf;            pend = buf + fread(buf, 1, BUF_SIZE, stdin);            if(pend == p1) {                IOerror = 1;                return -1;            }        }        return *p1++;    }    inline bool blank(char ch) {        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';    }    inline void read(int &x) {        char ch;        while(blank(ch = nc()));        if(IOerror)            return;        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');    }    #undef BUF_SIZE};using namespace fastIO;using namespace std;const int N = 3e6 + 5;struct p{    int l, r, id;   }a[N];ll mul[N];ll ans, tot;int pos;ll quickM(ll a, ll b){    ll ans = 1;    while(b){        if(b & 1)ans = ans * a % mo;        b >>= 1;        a = a * a % mo;    }    return ans;}int cmp(p a, p b){    return a.l < b.l || a.l == b.l && a.r > b.r;}ll C(ll n, ll m){    return mul[n] * quickM(mul[m], mo - 2) % mo * quickM(mul[n - m], mo - 2) % mo;  }void dfs(int l, int r){    int pos = tot;    if(r < l) return;    if(a[tot].l != l || a[tot].r != r) ans = 0;    if(ans == 0) return;    if(l == r) {        tot++;        return;    }    ans = ans * C(r - l, a[tot].id - l) % mo;    tot++;    dfs(l, a[pos].id - 1);    dfs(a[pos].id + 1, r);}int main(){    int n, kase = 0;    mul[0] = 1;    for(ll i = 1; i <= 1e6; i++) mul[i] = mul[i - 1] * i % mo;     while(read(n), !fastIO::IOerror){        ans = tot = 1;        for(int i = 1; i <= n; i++){            read(a[i].l);            a[i].id = i;        }        for(int i = 1; i <= n; i++){            read(a[i].r);        }        sort(a + 1, a + n + 1, cmp);        dfs(1, n);        printf("Case #%d: %lld\n", ++kase, ans);    }    return 0;}