HDU 6044 Limited Permutation ?????????+????????????
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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.
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.
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
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????????????pi?????? L ,R,???????????????????????????????????????pi???????????????
???????????????????????????pi ????????? 1???n???????????? ?????? pj???pk?????? ?????? ???1???i - 1?????? ???i+1,n???
??????dfs??????????????????
?????????????????????????????????????????????
#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????????????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????????????Finv?????????????????? 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)??????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\n",cas++,dfs(1,n)); } return 0; }
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