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Limited Permutation
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1714 Accepted Submission(s): 463
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
题意:给你n个l r 代表[li,ri]的最小值是pi且[li,ri+1]或[li-1,ri]的最小值不是pi 问能构造出多少个这样的序列
题解:对于当前区间 如果我们找不到有[l,r]为当前区间 那么方案数就是0
假设有[l,l+x],[l+x+1,r]两个区间
如果[l,l+x]是[l,r]这个区间的最小值 那么会并吞掉[l+x+1,r]
反之也成立
所以组合数搞一搞就ok。。。
#include <cstdio>
#include <algorithm>
#include <utility>
#include <map>
using namespace std;
typedef long long LL;
typedef pair<int,int> P;
const LL mod=1000000007;
const int N=1000010;
namespace fastIO{
#define BUF_SIZE 100000
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 bool read(int &x){
char ch;
while(blank(ch=nc()));
if(IOerror)return 0;
for(x=ch-'0';(ch=nc())>='0'&&ch<='9';x=x*10+ch-'0');
return 1;
}
#undef BUF_SIZE
};
namespace Comb{
int f[N],rf[N];
LL inv(LL a,LL m){return(a==1?1:inv(m%a,m)*(m-m/a)%m);}
LL C(int n,int m){
if(m<0||m>n)return 0;
return (LL)f[n]*rf[m]%mod*rf[n-m]%mod;
}
void init(){
f[0]=1;for(int i=1;i<=1000000;i++)f[i]=(LL)f[i-1]*i%mod;
rf[1000000]=inv(f[1000000],mod);
for(int i=1000000;i;i--)rf[i-1]=(LL)rf[i]*i%mod;
}
}
int l[N],r[N],n;
map<P,int> M;
LL dfs(int l,int r){
// printf("%d %d\n",l,r);
if(l>r)return 1;
int x=M[P(l,r)]; if(!x)return 0;
return Comb::C(r-l,r-x)*dfs(l,x-1)%mod*dfs(x+1,r)%mod;
}
int main(){
Comb::init();
for(int cas=1;fastIO::read(n);cas++){
M.clear();
for(int i=1;i<=n;i++)fastIO::read(l[i]);
for(int i=1;i<=n;i++)fastIO::read(r[i]);
for(int i=1;i<=n;i++)M[P(l[i],r[i])]=i;
LL ans=dfs(1,n);
printf("Case #%d: %lld\n",cas,ans);
}return 0;
}