HDU 6044--Limited Permutation(搜索+组合数+逆元)



Problem Description
As to a permutation p1,p2,,pn from 1 to n, it is uncomplicated for each 1in to calculate (li,ri) meeting the condition that min(pL,pL+1,,pR)=pi if and only if liLiRri for each 1LRn.

Given the positive integers n(li,ri) (1in), 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.


The input contains multiple test cases.

For each test case:

The first line contains one positive integer n, satisfying 1n106.

The second line contains n positive integers l1,l2,,ln, satisfying 1lii for each 1in.

The third line contains n positive integers r1,r2,,rn, satisfying irin for each 1in.

It's guaranteed that the sum of n in all test cases is not larger than 3106.

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.


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


Sample Input
1 1 3
1 3 3
1 2 2 4 5
5 2 5 5 5


Sample Output
Case #1: 2 
Case #2: 3
题意:对于一个由1~n组成的排列称为合法的,必须满足这样的条件: 有n个(li,ri),对于每个 i 必须满足 min(pL,pL+1,,pR)=pi if and only if liLiRri for each 1LRn.



思路:对于每个(li,ri),a[li-1]<a[i]>a[ri+1] ,并且a[i]是a[li]~a[ri]的最小值,那么可以想到:对于区间(1,n)一定有一个最小值,所以一定有一个区间是(1,n)(用X表示),那么这个最小值把区间X分成两部分U和V ,所以一定存在为U和V的区间,如果不存在,那么输出0,令f(X)表示符合条件的区间X的排列数,那么f(X)=f(U)*f(V)*C(U+V+1,U)   【注:C()表示组合数,U 表示区间U的大小】,所以我们只需要从区间(1,n)进行深搜即可,其中因为数据太大取模会用到逆元和组合数。



#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int N=1e6+5;
const LL mod=1e9+7;
LL fac[N], Inv[N];
/*int Scan()///输入外挂
    int res=0,ch,flag=0;
    else if(ch>='0'&&ch<='9')
    return flag?-res:res;
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;

void Init(){
    fac[0] = Inv[0] = fac[1] = Inv[1] = 1;
    for(int i=2; i<N; i++) fac[i] = fac[i-1] * i % mod;
    for(int i=2; i<N; i++) Inv[i] = (mod - mod / i) * Inv[mod % i] % mod;
    for(int i=2; i<N; i++) Inv[i] = Inv[i] * Inv[i-1] % mod;
LL C(LL n, LL m){
    LL ans = fac[n] * Inv[m] % mod* Inv[n-m] %mod;
    return ans;
int ii;
struct Node{
    int l,r;
    int id;
bool cmp(const Node s1,const Node s2)
    if(s1.l==s2.l) return s1.r>s2.r;
    return s1.l<s2.l;
LL dfs(int L,int R)
    if(a[ii].l!=L || a[ii].r!=R)  return 0;
    int m=a[ii++].id;
    LL fL=1,fR=1;
    if(L<=m-1) fL=dfs(L,m-1);
    if(m+1<=R) fR=dfs(m+1,R);
    LL c=C(R-L,m-L);
    return fL*fR%mod*c%mod;
int main()
    int n,Case=1;
       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;
       LL ans=dfs(1,n);
       printf("Case #%d: %lld\n",Case++,ans);
    return 0;


posted @ 2017-07-30 11:16  茶飘香~  阅读(...)  评论(...编辑  收藏